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##### Learning Objectives

- To balance equations that describe reactions in solution.
- To calculate the quantities of compounds produced or consumed in a chemical reaction.
- To solve quantitative problems involving the stoichiometry of reactions in solution.

A balanced chemical equation gives the identity of the reactants and the products as well as the accurate number of molecules or moles of each that are consumed or produced. **Stoichiometry **is a collective term for the quantitative relationships between the masses, the numbers of moles, and the numbers of particles (atoms, molecules, and ions) of the reactants and the products in a balanced chemical equation. **A stoichiometric quantity** is the amount of product or reactant specified by the coefficients in a balanced chemical equation. This section describes how to use the stoichiometry of a reaction to answer questions like the following: How much oxygen is needed to ensure complete combustion of a given amount of isooctane? (This information is crucial to the design of nonpolluting and efficient automobile engines.) How many grams of pure gold can be obtained from a ton of low-grade gold ore? (The answer determines whether the ore deposit is worth mining.) If an industrial plant must produce a certain number of tons of sulfuric acid per week, how much elemental sulfur must arrive by rail each week?

All these questions can be answered using the concepts of the mole, molar and formula masses, and solution concentrations, along with the coefficients in the appropriate balanced chemical equation.

## Stoichiometry Problems

When carrying out a reaction in either an industrial setting or a laboratory, it is easier to work with masses of substances than with the numbers of molecules or moles. The general method for converting from the mass of any reactant or product to the mass of any other reactant or product using a balanced chemical equation is outlined in and described in the following text.

## Steps in Converting between Masses of Reactant and Product

- Convert the mass of one substance (substance A) to the corresponding number of moles using its molar mass.
- From the balanced chemical equation, obtain the number of moles of another substance (B) from the number of moles of substance A using the appropriate mole ratio (the ratio of their coefficients).
- Convert the number of moles of substance B to mass using its molar mass. It is important to remember that some species are present in excess by virtue of the reaction conditions. For example, if a substance reacts with the oxygen in air, then oxygen is in obvious (but unstated) excess.

Converting amounts of substances to moles—and vice versa—is the key to all stoichiometry problems, whether the amounts are given in units of mass (grams or kilograms), weight (pounds or tons), or volume (liters or gallons).

To illustrate this procedure, consider the combustion of glucose. Glucose reacts with oxygen to produce carbon dioxide and water:

\[ C_6H_{12}O_6 (s) + 6 O_2 (g) \rightarrow 6 CO_2 (g) + 6 H_2O (l) \label{3.6.1} \]

Just before a chemistry exam, suppose a friend reminds you that glucose is the major fuel used by the human brain. You therefore decide to eat a candy bar to make sure that your brain does not run out of energy during the exam (even though there is no direct evidence that consumption of candy bars improves performance on chemistry exams). If a typical 2 oz candy bar contains the equivalent of 45.3 g of glucose and the glucose is completely converted to carbon dioxide during the exam, how many grams of carbon dioxide will you produce and exhale into the exam room?

The initial step in solving a problem of this type is to write the balanced chemical equation for the reaction. Inspection shows that it is balanced as written, so the strategy outlined above can be adapted as follows:

1. Use the molar mass of glucose (to one decimal place, 180.2 g/mol) to determine the number of moles of glucose in the candy bar:

\[ moles \, glucose = 45.3 \, g \, glucose \times {1 \, mol \, glucose \over 180.2 \, g \, glucose } = 0.251 \, mol \, glucose \nonumber \]

2. According to the balanced chemical equation, 6 mol of CO_{2} is produced per mole of glucose; the mole ratio of CO_{2} to glucose is therefore 6:1. The number of moles of CO_{2} produced is thus

\[ moles \, CO_2 = mol \, glucose \times {6 \, mol \, CO_2 \over 1 \, mol \, glucose } \nonumber \]

\[ = 0.251 \, mol \, glucose \times {6 \, mol \, CO_2 \over 1 \, mol \, glucose } \nonumber \]

\[ = 1.51 \, mol \, CO_2 \nonumber \]

3. Use the molar mass of CO_{2} (44.010 g/mol) to calculate the mass of CO_{2} corresponding to 1.51 mol of CO_{2}:

\[ mass\, of\, CO_2 = 1.51 \, mol \, CO_2 \times {44.010 \, g \, CO_2 \over 1 \, mol \, CO_2} = 66.5 \, g \, CO_2 \nonumber \]

These operations can be summarized as follows:

\[ 45.3 \, g \, glucose \times {1 \, mol \, glucose \over 180.2 \, g \, glucose} \times {6 \, mol \, CO_2 \over 1 \, mol \, glucose} \times {44.010 \, g \, CO_2 \over 1 \, mol \, CO_2} = 66.4 \, g \, CO_2 \nonumber \]

Discrepancies between the two values are attributed to rounding errors resulting from using stepwise calculations in steps 1–3. (Remember that you should generally carry extra significant digits through a multistep calculation to the end to avoid this!) This amount of gaseous carbon dioxide occupies an enormous volume—more than 33 L. Similar methods can be used to calculate the amount of oxygen consumed or the amount of water produced.

The balanced chemical equation was used to calculate the mass of product that is formed from a certain amount of reactant. It can also be used to determine the masses of reactants that are necessary to form a certain amount of product or, as shown in Example \(\PageIndex{1}\), the mass of one reactant that is required to consume a given mass of another reactant.

##### Example \(\PageIndex{1}\): The US Space Shuttle

The combustion of hydrogen with oxygen to produce gaseous water is extremely vigorous, producing one of the hottest flames known. Because so much energy is released for a given mass of hydrogen or oxygen, this reaction was used to fuel the NASA (National Aeronautics and Space Administration) space shuttles, which have recently been retired from service. NASA engineers calculated the exact amount of each reactant needed for the flight to make sure that the shuttles did not carry excess fuel into orbit. Calculate how many tons of hydrogen a space shuttle needed to carry for each 1.00 tn of oxygen (1 tn = 2000 lb).

* The US space shuttle Discovery during liftoff. The large cylinder in the middle contains the oxygen and hydrogen that fueled the shuttle’s main engine*.

**Given**: reactants, products, and mass of one reactant

**Asked for**: mass of other reactant

**Strategy**:

- Write the balanced chemical equation for the reaction.
- Convert mass of oxygen to moles. From the mole ratio in the balanced chemical equation, determine the number of moles of hydrogen required. Then convert the moles of hydrogen to the equivalent mass in tons.

**Solution**:

We use the same general strategy for solving stoichiometric calculations as in the preceding example. Because the amount of oxygen is given in tons rather than grams, however, we also need to convert tons to units of mass in grams. Another conversion is needed at the end to report the final answer in tons.

A We first use the information given to write a balanced chemical equation. Because we know the identity of both the reactants and the product, we can write the reaction as follows:

\[ H_2 (g) + O_2 (g) \rightarrow H_2O (g) \nonumber \]

This equation is not balanced because there are two oxygen atoms on the left side and only one on the right. Assigning a coefficient of 2 to both H_{2}O and H_{2} gives the balanced chemical equation:

\[ 2 H_2 (g) + O_2 (g) \rightarrow 2 H_2O (g) \nonumber \]

Thus 2 mol of H_{2} react with 1 mol of O_{2} to produce 2 mol of H_{2}O.

1. **B** To convert tons of oxygen to units of mass in grams, we multiply by the appropriate conversion factors:

\[ mass \, of \, O_2 = 1.00 \, tn \times { 2000 \, lb \over tn} \times {453.6 \, g \over lb} = 9.07 \times 10^5 \, g \, O_2 \nonumber \]

Using the molar mass of O_{2} (32.00 g/mol, to four significant figures), we can calculate the number of moles of O_{2} contained in this mass of O_{2}:

\[ mol \, O_2 = 9.07 \times 10^5 \, g \, O_2 \times {1 \, mol \, O_2 \over 32.00 \, g \, O_2} = 2.83 \times 10^4 \, mol \, O_2 \nonumber \]

2. Now use the coefficients in the balanced chemical equation to obtain the number of moles of H_{2} needed to react with this number of moles of O_{2}:

\[ mol \, H_2 = mol \, O_2 \times {2 \, mol \, H_2 \over 1 \, mol \, O_2} \nonumber \]

\[ = 2.83 \times 10^4 \, mol \, O_2 \times {2 \, mol \, H_2 \over 1 \, mol \, O_2} = 5.66 \times 10^4 \, mol \, H_2 \nonumber \]

3. The molar mass of H_{2} (2.016 g/mol) allows us to calculate the corresponding mass of H_{2}:

\[mass \, of \, H_2 = 5.66 \times 10^4 \, mol \, H_2 \times {2.016 \, g \, H_2 \over mol \, H_2} = 1.14 \times 10^5 \, g \, H_2 \nonumber \]

Finally, convert the mass of H2 to the desired units (tons) by using the appropriate conversion factors:

\[ tons \, H_2 = 1.14 \times 10^5 \, g \, H_2 \times {1 \, lb \over 453.6 \, g} \times {1 \, tn \over 2000 \, lb} = 0.126 \, tn \, H_2 \nonumber \]

The space shuttle had to be designed to carry 0.126 tn of H_{2} for each 1.00 tn of O_{2}. Even though 2 mol of H_{2} are needed to react with each mole of O_{2}, the molar mass of H_{2} is so much smaller than that of O_{2} that only a relatively small mass of H_{2} is needed compared to the mass of O_{2}.

##### Exercise \(\PageIndex{1}\): Roasting Cinnabar

Cinnabar, (or Cinnabarite) \(HgS\) is the common ore of mercury. Because of its mercury content, cinnabar can be toxic to human beings; however, because of its red color, it has also been used since ancient times as a pigment.

Alchemists produced elemental mercury by roasting cinnabar ore in air:

\[ HgS (s) + O_2 (g) \rightarrow Hg (l) + SO_2 (g) \nonumber \]

The volatility and toxicity of mercury make this a hazardous procedure, which likely shortened the life span of many alchemists. Given 100 g of cinnabar, how much elemental mercury can be produced from this reaction?

**Answer**-
86.2 g

## Calculating Moles from Volume

Quantitative calculations involving reactions in solution are carried out with *masses*, however, *volumes* of solutions of known concentration are used to determine the number of moles of reactants. Whether dealing with volumes of solutions of reactants or masses of reactants, the coefficients in the balanced chemical equation give the number of moles of each reactant needed and the number of moles of each product that can be produced. An expanded version of the flowchart for stoichiometric calculations is shown in Figure \(\PageIndex{2}\). The balanced chemical equation for the reaction and *either* the masses of solid reactants and products *or* the volumes of solutions of reactants and products can be used to determine the amounts of other species, as illustrated in the following examples.

The balanced chemical equation for a reaction and either the masses of solid reactants and products or the volumes of solutions of reactants and products can be used in stoichiometric calculations.

##### Example \(\PageIndex{2}\) : Extraction of Gold

Gold is extracted from its ores by treatment with an aqueous cyanide solution, which causes a reaction that forms the soluble [Au(CN)_{2}]^{−} ion. Gold is then recovered by reduction with metallic zinc according to the following equation:

\[ Zn(s) + 2[Au(CN)_2]^-(aq) \rightarrow [Zn(CN)_4]^{2-}(aq) + 2Au(s) \nonumber \]

What mass of gold can be recovered from 400.0 L of a 3.30 × 10^{−4} M solution of [Au(CN)_{2}]^{−}?

**Given: **chemical equation and molarity and volume of reactant

**Asked for: **mass of product

###### Strategy:

- Check the chemical equation to make sure it is balanced as written; balance if necessary. Then calculate the number of moles of [Au(CN)
_{2}]^{−}present by multiplying the volume of the solution by its concentration. - From the balanced chemical equation, use a mole ratio to calculate the number of moles of gold that can be obtained from the reaction. To calculate the mass of gold recovered, multiply the number of moles of gold by its molar mass.

###### Solution:

**A** The equation is balanced as written; proceed to the stoichiometric calculation. Figure \(\PageIndex{2}\) is adapted for this particular problem as follows:

As indicated in the strategy, start by calculating the number of moles of [Au(CN)_{2}]^{−} present in the solution from the volume and concentration of the [Au(CN)_{2}]^{−} solution:

\( \begin{align} moles\: [Au(CN)_2 ]^-

& = V_L M_{mol/L} \\

& = 400 .0\: \cancel{L} \left( \dfrac{3 .30 \times 10^{4-}\: mol\: [Au(CN)_2 ]^-} {1\: \cancel{L}} \right) = 0 .132\: mol\: [Au(CN)_2 ]^- \end{align} \)

**B** Because the coefficients of gold and the [Au(CN)_{2}]^{−} ion are the same in the balanced chemical equation, assuming that Zn(s) is present in excess, the number of moles of gold produced is the same as the number of moles of [Au(CN)_{2}]^{−} (i.e., 0.132 mol of Au). The problem asks for the mass of gold that can be obtained, so the number of moles of gold must be converted to the corresponding mass using the molar mass of gold:

\( \begin{align} mass\: of\: Au &= (moles\: Au)(molar\: mass\: Au) \\

&= 0 .132\: \cancel{mol\: Au} \left( \dfrac{196 .97\: g\: Au} {1\: \cancel{mol\: Au}} \right) = 26 .0\: g\: Au \end{align}\)

At a 2011 market price of over $1400 per troy ounce (31.10 g), this amount of gold is worth $1170.

\( 26 .0\: \cancel{g\: Au} \times \dfrac{1\: \cancel{troy\: oz}} {31 .10\: \cancel{g}} \times \dfrac{\$1400} {1\: \cancel{troy\: oz\: Au}} = \$1170 \)

##### Exercise \(\PageIndex{2}\) : Lanthanum Oxalate

What mass of solid lanthanum(III) oxalate nonahydrate [La_{2}(C_{2}O_{4})_{3}•9H_{2}O] can be obtained from 650 mL of a 0.0170 M aqueous solution of LaCl_{3} by adding a stoichiometric amount of sodium oxalate?

**Answer**-
3.89 g

Finding Mols and Masses of Reactants and Products Using Stoichiometric Factors (Mol Ratios): Finding Mols and Masses of Reactants and Products Using Stoichiometric Factors, YouTube(opens in new window) [youtu.be]

## Summary

Either the masses or the volumes of solutions of reactants and products can be used to determine the amounts of other species in the balanced chemical equation. Quantitative calculations that involve the stoichiometry of reactions in solution use volumes of solutions of known concentration instead of masses of reactants or products. The coefficients in the balanced chemical equation tell how many moles of reactants are needed and how many moles of product can be produced.

## FAQs

### What is Step 5 in stoichiometry? ›

Determining the stoichiometry of species. 5. **Calculating the desired quantity**.

**What are the 4 types of stoichiometry problems? ›**

Stoichiometric difficulties can be classified into four categories: **Mass to mass conversion**. Mass to moles conversion. Mole to mass steps conversion.

**How do you solve stoichiometry solutions easy? ›**

**5 Simple Steps to Solve Solution Stoichiometry Problems**

- Figure out if it's an M = n/V problem or a McVc = MdVd problem. M = n/V. ...
- See what you need to solve for. ...
- For M=n/V, convert everything to moles. ...
- For McVc=MdVd, solve for 3 of the 4 variables. ...
- For “How would you prepare X solution” problems…

**What are the 3 types of stoichiometry problems? ›**

**There are three types of Gas Stoichiometry problems:**

- Mole-Volume (or Volume-Mole)
- Mass-volume (or volume-mass)
- Volume-Volume.

**What is stoichiometry example? ›**

Example – Using Stoichiometric Ratio (Moles)

By looking at the coefficients, you can see that **for every 1 mole of C _{6}H_{12}O_{6}, 2 moles of CO_{2} are produced**. Using this ratio, you can figure out how many moles of carbon dioxide are made from 2.5 moles of glucose.

**What is the formula for mass stoichiometry? ›**

The main equation is: **moles A x (mole ratio of B/A) x molar mass of B = mass of B**. The way you would use this in an actual problem is the following. You are asked how many grams of glucose are produced when 3 moles of water react with carbon dioxide. Start with your balanced equation: 6 CO2 + 6 H2 O → C6 H12 O6 + 6 O2.

**What is stoichiometry solution example? ›**

Solution Stoichiometry Movie Text

It is defined as the moles of a substance contained in one liter of solution. For instance, **if a solution has a concentration of 1.20 M NaCl, this means that there are 1.20 moles of NaCl per liter of solution**.

**How do you calculate mole ratio? ›**

Convert the mass of each element to moles of each element using the atomic masses. Find the ratio or the moles of each element by **dividing the number of moles of each by the smallest number of moles**.

**What is stoichiometry in math? ›**

Stoichiometry is **used to converting from units from grams to moles**. This conversion is required because atoms and molecules are too small to count in a meaningful way so we use molar mass find the number of moles of atoms or molecules. This number can then be used in a chemical equation.

**Is stoichiometry the hardest part of chemistry? ›**

**Stoichiometry is arguably one of the most difficult concepts for students to grasp in a general chemistry class**. Stoichiometry requires students to synthesize their knowledge of moles, balanced equations and proportional reasoning to describe a process that is too small to see.

### What is the first step in most stoichiometry? ›

the first step in any stoichiometric problem is to **always ensure that the chemical reaction you are dealing with is balanced**, clarity of the concept of a 'mole' and the relationship between 'amount (grams)' and 'moles'.

**How difficult is stoichiometry? ›**

Stoichiometry can be difficult because it builds upon a number of individual skills. To be successful you must master the skills and learn how to plan your problem solving strategy. Master each of these skills before moving on: Calculating Molar Mass.

**How do you calculate reactants? ›**

**Worked Example of Using Mole Ratio to Calculate Mass of Reactant or Product**

- mass O
_{2}= moles(O_{2}) × molar mass(O_{2}) (a) Calculate moles(Mg) = mass(Mg) ÷ molar mass(Mg) moles(Mg) = 12.2 ÷ 24.31 = 0.50 mol. ... - mass MgO = moles(MgO) × molar mass(MgO) (a) Calculate moles Mg. moles(Mg) = mass(Mg) ÷ molar mass(Mg)

**What is ideal stoichiometric calculations? ›**

In an ideal stoichiometry problem, **the mass of any reactant or product can be calculated from the mass of any other reactant or product if the balanced equation is known**. Since the molar volume of any gas at STP is constant, gas volumes can also be used in stoichiometric calculations.

**What is the first step in all stoichiometric calculations? ›**

The first and critical step in any stoichiometric calculation is to **have a balanced chemical equation**.

**What is stoichiometry in a nutshell? ›**

The best definition for stoichiometry is the simple one: **it's a way to figure out how much stuff you're going to make in a chemical reaction, or how much stuff you'll need to make a chemical reaction do what you want**.

**Is stoichiometry taught in high school? ›**

Stoichiometry, "the quantitative relationship between two or more substances especially in processes involving physical or chemical change" (Merriam-Webster), is **currently a major part of the U.S. high school curriculum**.

**What type of math is stoichiometry? ›**

Stoichiometry is simply **the math behind chemistry**. Given enough information, one can use stoichiometry to calculate masses, moles, and percents within a chemical equation.

**What are 2 examples of the application of stoichiometry? ›**

Stoichiometry is considered the heart of chemistry with so many practical uses. The composition of all the chemical products we use in our daily lives, such as **shampoos, cleaners, perfumes, soaps, and fertilizers** is formed using stoichiometric calculations.

**What are 5 real life application of stoichiometry? ›**

**Soap, tires, fertilizer, gasoline, deodorant, and chocolate bars** are just a few commodities you use that are chemically engineered, or produced through chemical reactions. Chemically engineered commodities all rely on stoichiometry for their production.

### How do we use stoichiometry in chemistry? ›

Stoichiometry uses the proportional nature of chemical equations **to determine the amount of reactant needed to produce a given amount of product or predict the amount that will be produced from a given amount of reactant**.

**How is a mole ratio used in stoichiometry? ›**

It uses the mole ratios between the different parts of the reaction **to calculate how much of one substance will be needed to produce a certain amount of another substance**. Or another way to look at it is the mole ratio is used to convert the number of moles of one substance to the number of moles of another substance.

**How do you stoichiometry moles and mass? ›**

The moles of the given substance are first converted into moles of the unknown by using the mole ratio from the balanced chemical equation. Then, the moles of the unknown are converted into mass in grams by use of the molar mass of that substance from the periodic table.

**What is a ratio in stoichiometry? ›**

stoichiometric ratio: **The ratio of the coefficients of the products and reactants in a balanced reaction**. This ratio can be used to calculate the amount of products or reactants produced or used in a reaction.

**What is a mole ratio example? ›**

Mole Ratio Example: Balanced Equation

The mole ratio between O_{2} and H_{2}O is 1:2. For every 1 mole of O_{2} used, 2 moles of H_{2}O are formed. The mole ratio between H_{2} and H_{2}O is 1:1. For every 2 moles of H_{2} used, 2 moles of H_{2}O are formed.

**Is stoichiometry a mole? ›**

**A balanced chemical reaction gives equivalences in moles that allow stoichiometry calculations to be performed**. Mole quantities of one substance can be related to mass quantities using a balanced chemical equation. Mass quantities of one substance can be related to mass quantities using a balanced chemical equation.

**Is stoichiometry the same as moles? ›**

Stoichiometry is exactly that. **It is the quantitative relation between the number of moles (and therefore mass) of various products and reactants in a chemical reaction**. Chemical reactions must be balanced, or in other words, must have the same number of various atoms in the products as in the reactants.

**Why do we use stoichiometry? ›**

Stoichiometry **gives us the quantitative tools to figure out the relative amounts of reactants and products in chemical reactions**.

**Why is high school chemistry so hard? ›**

The primary reason chemistry is so hard is **because of the topic progression**. You really have to fully understand several topics before you can fully understand other topics. It's important to keep in mind, memorization isn't the key here. There's a certain element of memorization.

**What grade do you learn stoichiometry? ›**

Stoichiometry concepts taught at secondary level, mostly for **grade 11 and grade 12**, involve problem-solving and under standing of chemical reactions and equations.

### What is high school chemistry like? ›

High school chemistry **introduces students to more complicated topics such as stoichiometry, thermodynamics, virtual laboratory experiments, and more**. This strong base of knowledge will prepare them for subjects such as physics and advanced biology in the near future.

**Why is it called stoichiometry? ›**

Stoichiometry is the study of the quantitative, or measureable, relationships that exist in chemical formulas and chemical reactions. The term is **derived from the Greek words stoicheion, meaning element, and metron, meaing measure**.

**What do all stoichiometric calculations begin with? ›**

All stoichiometric calculations begin with **a balanced equation**. Balanced equations are necessary because mass is conserved in every reaction. The number and kinds of atoms must be the same in the reactants and products.

**How do you calculate gas stoichiometry? ›**

To account for these conditions, we use the ideal gas equation **PV=nRT** where P is the pressure measured in atmosphere(atm), V is the volume measured in liters (L), n is the number of moles, R is the gas constant with a value of . 08206 L atm mol^{-}^{1} K^{-}^{1}, and T is the temperature measured in kelvin (K).

**What is the stoichiometric method? ›**

Stoichiometry is a section of chemistry that involves **using relationships between reactants and/or products in a chemical reaction to determine desired quantitative data**. In Greek, stoikhein means element and metron means measure, so stoichiometry literally translated means the measure of elements.

**How do nurses use stoichiometric calculations? ›**

Nurses routinely use stoichiometry **to convert the drip ratios of medication to the prescribed dosages**. They use a persons weight and conversion factors to determine the correct dosage of medicine to give. If the stoichiometry calculations are done incorrectly the patient can potentially be harmed.

**Why is stoichiometry so hard? ›**

Stoichiometry can be difficult because **it builds upon a number of individual skills**. To be successful you must master the skills and learn how to plan your problem solving strategy. Master each of these skills before moving on: Calculating Molar Mass.

**What are the numbers in stoichiometry? ›**

Stoichiometric coefficient or stoichiometric number is **the number of molecules that participate in the reaction**. If you look at any balanced reaction you will notice that there are an equal number of elements on both sides of the equation.

**How do you find volume using stoichiometry? ›**

**Here are the steps that one would take:**

- Convert mass of given to moles of given by dividing by the formula mass of the given.
- Convert moles of given to moles of needed using the balanced equation.
- Convert moles of needed to liters of requested using PV = nRT.

**What is the formula to calculate volume of gas? ›**

First, let's review the ideal gas law, **PV = nRT**. In this equation, 'P' is the pressure in atmospheres, 'V' is the volume in liters, 'n' is the number of particles in moles, 'T' is the temperature in Kelvin and 'R' is the ideal gas constant (0.0821 liter atmospheres per moles Kelvin).

### What is the first step in solving stoichiometry problem? ›

The first and critical step in any stoichiometric calculation is to **have a balanced chemical equation**.

**What is the first and most important step in stoichiometric computation? ›**

the first step in any stoichiometric problem is to **always ensure that the chemical reaction you are dealing with is balanced**, clarity of the concept of a 'mole' and the relationship between 'amount (grams)' and 'moles'.

**What is the correct stoichiometric ratio? ›**

The stoichiometric mixture for a gasoline engine is the ideal ratio of air to fuel that burns all fuel with no excess air. For gasoline fuel, the stoichiometric air–fuel mixture is about **14.7:1** i.e. for every one gram of fuel, 14.7 grams of air are required.

**Is there a lot of math in nursing? ›**

Nursing in the "real world" **generally requires very basic math skills, but almost all programs require at least one college-level math class — usually algebra**. Some nursing schools may require a basic statistics course as well, so if you know what schools you're applying to, be sure to check for this requirement.