Question

stoichiometry & percent yield worksheet
show all work!
% yield = \\( \frac { \text { actual yield } } { \text { theoretical yield } } \times 100 \\)
theoretical yield = answer to your stoich problem
actual yield = given in the problem or the experimental yield

1. balance the equation for the reaction of iron (iii) phosphate with sodium sulfate to make iron (iii) sulfate and sodium phosphate.
2. if i perform this reaction with 25g of iron (iii) phosphate and an excess of sodium sulfate, how many grams of iron (iii) sulfate can i make?
3. if 18.5g of iron (iii) sulfate are actually made when i do this reaction, what is my percent yield?
4. is the answer from problem #3 reasonable? explain.
5. \\( \mathrm { lioh } + \mathrm { kcl } \

ightarrow \mathrm { licl } + \mathrm { koh } \\)
a. i began this reaction with 20.0g of lithium hydroxide. what is my theoretical yield of lithium chloride?
b. i actually produced 6.0g of lithium chloride. what is my percent yield?

1. \\( \mathrm { c } _ { 3 } \mathrm { h } _ { 8 } + 5 \mathrm { o } _ { 2 } \

ightarrow 3 \mathrm { co } _ { 2 } + 4 \mathrm { h } _ { 2 } \mathrm { o } \\)
a. if i start with 5g of \\( \mathrm { c } _ { 3 } \mathrm { h } _ { 8 } \\), what is my theoretical yield of water?
b. i got a percent yield of 75%. how many grams of water did i make?

1. \\( \mathrm { be } + 2 \mathrm { hcl } \

ightarrow \mathrm { becl } _ { 2 } + \mathrm { h } _ { 2 } \\)
a. my theoretical yield of beryllium chloride was 10.7g. if my actual yield was 4.5g, what was the percent yield?

Explanation:

Step1: Balance the equation

The balanced chemical equation for the reaction of iron (III) phosphate ($\ce{FePO4}$) with sodium sulfate ($\ce{Na2SO4}$) is:

$$\ce{2FePO4 + 3Na2SO4 ightarrow Fe2(SO4)3 + 2Na3PO4}$$

Step2: Calculate moles of iron (III) phosphate

The molar mass of $\ce{FePO4}$ is $M = 55.85 + 30.97+4\times16=150.82\ \text{g/mol}$.
The number of moles of $\ce{FePO4}$, $n=\frac{m}{M}=\frac{25\ \text{g}}{150.82\ \text{g/mol}}\approx0.166\ \text{mol}$

Step3: Use stoichiometry to find moles of iron (III) sulfate

From the balanced equation, the mole ratio of $\ce{FePO4}$ to $\ce{Fe2(SO4)3}$ is $2:1$.
So, moles of $\ce{Fe2(SO4)3}$, $n_{\ce{Fe2(SO4)3}}=\frac{1}{2}\times n_{\ce{FePO4}}=\frac{1}{2}\times0.166\ \text{mol}= 0.083\ \text{mol}$

Step4: Calculate mass of iron (III) sulfate

The molar mass of $\ce{Fe2(SO4)3}$ is $M = 2\times55.85+3\times(32.07 + 4\times16)=399.88\ \text{g/mol}$
Mass of $\ce{Fe2(SO4)3}$, $m = n\times M=0.083\ \text{mol}\times399.88\ \text{g/mol}\approx33.2\ \text{g}$

Step5: Calculate percent yield

Given actual yield $m_{\text{actual}} = 18.5\ \text{g}$, theoretical yield $m_{\text{theoretical}}\approx33.2\ \text{g}$
Percent yield $=\frac{m_{\text{actual}}}{m_{\text{theoretical}}}\times100=\frac{18.5\ \text{g}}{33.2\ \text{g}}\times100\approx55.7\%$

Answer:

1. $\ce{2FePO4 + 3Na2SO4

ightarrow Fe2(SO4)3 + 2Na3PO4}$

1. Approximately $33.2\ \text{g}$
2. Approximately $55.7\%$
3. The percent yield of approximately $55.7\%$ is reasonable. In chemical reactions, there are often factors such as incomplete reactions, side - reactions, and losses during purification that can lead to yields less than $100\%$.
4. a. First, find moles of $\ce{LiOH}$: molar mass of $\ce{LiOH}=23.95\ \text{g/mol}$, $n_{\ce{LiOH}}=\frac{20.0\ \text{g}}{23.95\ \text{g/mol}}\approx0.835\ \text{mol}$. From the balanced equation $\ce{LiOH + KCl

ightarrow LiCl + KOH}$ (mole ratio $1:1$), moles of $\ce{LiCl}=n_{\ce{LiOH}}\approx0.835\ \text{mol}$. Molar mass of $\ce{LiCl}=42.39\ \text{g/mol}$, mass of $\ce{LiCl}=0.835\ \text{mol}\times42.39\ \text{g/mol}\approx35.4\ \text{g}$
b. Percent yield $=\frac{6.0\ \text{g}}{35.4\ \text{g}}\times100\approx16.9\%$

1. a. Molar mass of $\ce{C3H8}=44.1\ \text{g/mol}$, moles of $\ce{C3H8}=\frac{5\ \text{g}}{44.1\ \text{g/mol}}\approx0.113\ \text{mol}$. From the balanced equation $\ce{C3H8 + 5O2

ightarrow 3CO2 + 4H2O}$ (mole ratio $\ce{C3H8}: \ce{H2O}=1:4$), moles of $\ce{H2O}=4\times0.113\ \text{mol}=0.452\ \text{mol}$. Molar mass of $\ce{H2O}=18.02\ \text{g/mol}$, mass of $\ce{H2O}=0.452\ \text{mol}\times18.02\ \text{g/mol}\approx8.15\ \text{g}$
b. If percent yield $ = 75\%$ and theoretical yield $m_{\text{theoretical}}\approx8.15\ \text{g}$, actual yield $m_{\text{actual}}=0.75\times8.15\ \text{g}\approx6.11\ \text{g}$

1. a. Percent yield $=\frac{4.5\ \text{g}}{10.7\ \text{g}}\times100\approx42.1\%$