 ## Stoichiometry - exercises

$\require{mhchem}$

#### Worked Example 1

When $$64.5\textrm{g}$$ of methane is burnt in the presence of excess oxygen, it forms $$136.2\textrm{g}$$ of carbon dioxide and water vapour, as displayed in the following. However, the theoretical yield of carbon dioxide for this reaction is $$176.4\textrm{g}$$. Calculate the percentage yield of carbon dioxide for this reaction: $\ce{CH}_{4}\left(g\right)+2\ce{O}_{2}\left(g\right)\rightarrow\ce{CO}_{2}\left(g\right)+2\ce{H}_{2}\ce{O}\left(g\right)$

We are given actual and theoretical yields of carbon dioxide for this reaction and are asked to calculate the percentage yield of carbon dioxide for this reaction. We can use the following equation to find the percentage yield of carbon dioxide \begin{align*} \frac{\textrm{Actual yield}}{\textrm{Theoretical yield}}\times100 & =\frac{136.2\textrm{g}}{176.4\textrm{g}}\times100\\ = & 77.21\% \end{align*}

#### Worked Example 2

When $$168.3\textrm{g}$$ of $$\ce{Fe}$$ reacts with $$\ce{O}_{2}$$ according to the following chemical equation, $$172.6\textrm{g}$$ of $$\ce{Fe}_{2}\ce{O}_{3}$$ are obtained. What is the percentage yield of $$\ce{Fe}_{2}\ce{O}_{3}$$ for this reaction? Molar mass of $$\ce{Fe}_{2}\ce{O}_{3}$$ is $$159.69\textrm{g}/\textrm{mol}$$ and $$\ce{Fe}$$ is $$55.85\textrm{g}/\textrm{mol}$$. $4\ce{Fe}+3\ce{O}_{2}\rightarrow2\ce{Fe}_{2}\ce{O}_{3}$

We are given the mass of reactant $$\ce{Fe}$$, the actual yield of product $$\ce{Fe}_{2}\ce{O}_{3}$$, and the balanced chemical equation and are asked to calculate the percentage yield of $$\ce{Fe}_{2}\ce{O}_{3}$$. We need to find out the theoretical yield of $$\ce{Fe}_{2}\ce{O}_{3}$$ to calculate the percentage yield.

Step 1 - Use the molar ratios given in the balanced chemical equation and the given mass of $$\ce{Fe}$$ to calculate the theoretical yield of $$\ce{Fe}_{2}\ce{O}_{3}$$.

Number of moles of $$\ce{Fe}$$ \begin{align*} M & =\frac{m}{n}\\ n & =\frac{m}{M}\\ n & =\frac{168.3\textrm{g}}{55.85\textrm{g}/\textrm{mol}}\\ n & =3.013\textrm{mol} \end{align*}

From the balanced chemical equation, we know four moles of $$\ce{Fe}$$ forms two moles of $$\ce{Fe}_{2}\ce{O}_{3}$$. Therefore, $$3.013\textrm{mol}$$ of $$\ce{Fe}$$ forms the following amount of $$\ce{Fe}_{2}\ce{O}_{3}$$. \begin{align*} \frac{2\textrm{mol}}{4\textrm{mol}}\times3.013\textrm{mol} & =1.507\textrm{mol} \end{align*}

Convert the number of moles to $$\ce{Fe}_{2}\ce{O}_{3}$$ mass, \begin{align*} M & =\frac{m}{n}\\ m & =M\times n\\ m & =159.69\textrm{g}/\textrm{mol}\times1.507\textrm{mol}\\ m & =240.7\textrm{g} \end{align*}

Step 2 - Calculate the percentage yield of $$\ce{Fe}_{2}\ce{O}_{3}$$ using the calculated theoretical yield and the actual yield. \begin{align*} \frac{\textrm{Actual yield}}{\textrm{Theoretical yield}}\times100 & =\frac{172.6\textrm{g}}{240.7\textrm{g}}\times100=71.71\% \end{align*}

#### Exercises

1. $$\ce{CO}$$ reacts with $$\ce{O}_{2}$$ to form $$\ce{CO}_{2}$$ gas as shown below: $2\ce{CO}+\ce{O}_{2}\rightarrow2\ce{CO}_{2}$
1. How many moles of $$\ce{O}_{2}$$ are required to react with $$0.58\textrm{mol}$$ of $$\ce{CO}$$?
2. How many moles of $$\ce{CO}_{2}$$ can be produced when $$0.35\textrm{mol}$$ of $$\ce{CO}$$ reacts with $$\ce{O}_{2}$$?
3. How many moles of $$\ce{O}_{2}$$ must react with $$\ce{CO}$$ to produce $$0.48\textrm{mol}$$ of $$\ce{CO}_{2}$$?

a -$$0.29\textrm{mol}$$,

b -$$0.35\textrm{mol}$$,

c - $$0.24\textrm{mol}$$

1. The reaction between hydrazine and oxygen gas produce nitric oxide and water vapour according to the following balanced chemical equation: $\ce{N}_{2}\ce{H}_{4}\left(l\right)+3\ce{O}_{2}\left(g\right)\rightarrow2\ce{NO}_{2}\left(g\right)+2\ce{H}_{2}\ce{O}\left(g\right)$
1. How many moles of $$\ce{O}_{2}$$ are required to react with $$0.6\textrm{mol}$$ of $$\ce{N}_{2}\ce{H}_{4}$$?
2. How many moles of $$\ce{N}_{2}\ce{H}_{2}$$ must react with $$\ce{O}_{2}$$ to form $$0.4\textrm{mol}$$ of $$\ce{NO}_{2}$$?

a -$$2\textrm{mol}$$,

b -$$0.2\textrm{mol}$$

1. How many grams of $$\ce{O}_{2}$$ are required to form $$10.50\textrm{g}$$ of $$\ce{CO}_{2}$$ from the reaction between methane and oxygen as given below? The molar masses of $$\ce{CO}_{2}$$ and $$\ce{O}_{2}$$ are $$44.01\textrm{g}/\textrm{mol}$$ and $$32.00\textrm{g}/\textrm{mol}$$.
$$\ce{CH}_{4}+2\ce{O}_{2}\rightarrow\ce{CO}_{2}+2\ce{H}_{2}\ce{O}$$

$$15.27\textrm{g}$$

1. According to the following equation, how many grams of $$\ce{N}_{2}$$ are required to produce $$48.5\textrm{g}$$ of $$\ce{NH}_{3}$$? The molar masses of $$\ce{NH}_{3}$$ and $$\ce{N}_{2}$$ are $$17.03\textrm{g}/\textrm{mol}$$ and $$28.02\textrm{g}/\textrm{mol}$$. $3\ce{H}_{2}\left(g\right)+\ce{N}_{2}\left(g\right)\rightarrow2\ce{NH}_{3}\left(g\right)$

$$39.9\textrm{g}$$

1. Sulphur dioxide gas is produced by the reaction between sulphur and oxygen as follows: $\ce{S}+\ce{O}_{2}\rightarrow\ce{SO}_{2}$ If $$12.8\textrm{g}$$ of $$\ce{S}$$ reacts with $$18.5\textrm{g}$$ of $$\ce{O}_{2}$$;

1. Which one is the limiting reagent? The molar mass of $$\ce{S}$$ and $$\ce{O}_{2}$$ are $$32.07\textrm{g}/\textrm{mol}$$ and $$32.00\textrm{g}/\textrm{mol}$$.
2. What is the maximum mass of $$\ce{SO}_{2}$$ that can be formed under given conditions? The molar mass of $$\ce{SO}_{2}$$ is $$64.07\textrm{g}/\textrm{mol}$$.

a - $$\ce{S}$$,

b $$25.63\textrm{g}$$

1. $$\ce{HCl}$$ is formed according to the following reaction. $\ce{H}_{2}+\ce{Cl}_{2}\rightarrow2\ce{HCl}$ What is the maximum mass of $$\ce{HCl}$$ that can be produced when $$0.85\textrm{g}$$ of $$\ce{H}_{2}$$ reacts with $$42.58\textrm{g}$$ of $$\ce{Cl}_{2}$$? The atomic molar masses of $$\ce{H}$$ and $$\ce{Cl}$$ are $$1.01\textrm{g}/\textrm{mol}$$ and $$35.45\textrm{g}/\textrm{mol}$$.

$$31\textrm{g}$$

1. Calcium nitrite ($$\ce{Ca}_{3}\ce{N}_{2}$$) is produced using $$\ce{Ca}$$ and $$\ce{N}_{2}$$ as follows.
$3\ce{Ca}+\ce{N}_{2}\rightarrow\ce{Ca}_{3}\ce{N}_{2}$ What is the theoretical yield of $$\ce{Ca}_{3}\ce{N}_{2}$$ when $$120.56\textrm{g}$$ of $$\ce{Ca}$$ is reacted with $$42.87\textrm{g}$$ of $$\ce{N}_{2}$$? The atomic molar mass of $$\ce{Ca}$$ is $$40.08\textrm{g}/\textrm{mol}$$, and the atomic molar mass of $$\ce{N}$$ is $$14.01\textrm{g}/\textrm{mol}$$. The molar mass of $$\ce{Ca}_{3}\ce{N}_{2}$$ is $$148.26\textrm{g}/\textrm{mol}$$.

$$148.66\textrm{g}$$
1. Salicylic acid ($$\ce{C}_{7}\ce{H}_{6}\ce{O}_{3}$$) reacts with acetic anhydride ($$\ce{C}_{4}\ce{H}_{6}\ce{O}_{3}$$) to form acetylsalicylic acid ($$\ce{C}_{9}\ce{H}_{8}\ce{O}_{4}$$), which we call aspirin. The percentage yield of aspirin from the following reaction is $$88.48\%$$. $2\ce{C}_{7}\ce{H}_{6}\ce{O}_{3}+\ce{C}_{4}\ce{H}_{6}\ce{O}_{3}\rightarrow2\ce{C}_{9}\ce{H}_{8}\ce{O}_{4}+\ce{H}_{2}\ce{O}$ How many grams of aspirin were actually formed by the reaction of $$250.82\textrm{g}$$ of salicylic acid? The molar masses of salicylic acid and acetylsalicylate acid are $$138.21\textrm{g}/\textrm{mol}$$ and $$180.16\textrm{g}$$, respectively.
$$289.3\textrm{g}$$