Question

1. what is the percent yield of $\text{co}_2(g)$ if 46.5 grams of $\text{co}_2(g)$ is recovered from the reaction of 21.5 grams of $\text{c}_2\text{h}_2(g)$ according to the reaction below?

$2\text{c}_2\text{h}_2(g) + 5\text{o}_2(g) \
ightarrow 4\text{co}_2(g) + 2\text{h}_2\text{o}(g)$

1. what is the percent yield below if 3.12 grams of $\text{fe}(s)$ is recovered from a reaction whose theoretical yield of $\text{fe}(s)$ is 4.51 grams?

$\text{fe}_2\text{o}_3(s) + 3\text{co}(g) \
ightarrow 2\text{fe}(s) + 3\text{co}_2(g)$

1. what is the theoretical yield, in grams, of $\text{lihco}_3(s)$ that can be produced from 45.3 grams of $\text{lioh}(s)$ according to the reaction below?

$\text{lioh}(s) + \text{co}_2(g) \
ightarrow \text{lihco}_3(s)$

1. the percent yield for a reaction is 51.4%. the actual amount of product produced was 72.8 grams. what was the theoretical yield, in grams?
2. what is the percent yield if 14.0 grams of $\text{nacl}(s)$ is recovered from the reaction of 9.16 grams of $\text{cl}_2(g)$ according to the reaction below?

$2\text{na}(s) + \text{cl}_2(g) \
ightarrow 2\text{nacl}(s)$

Explanation:

(Question 6):

Step1: Molar mass of $\text{C}_2\text{H}_2$

Molar mass of $\text{C}_2\text{H}_2 = (2\times12.01)+(2\times1.008) = 26.036\ \text{g/mol}$

Step2: Moles of $\text{C}_2\text{H}_2$

$\text{Moles of }\text{C}_2\text{H}_2 = \frac{21.5\ \text{g}}{26.036\ \text{g/mol}} \approx 0.826\ \text{mol}$

Step3: Mole ratio to $\text{CO}_2$

From reaction: $2\ \text{mol }\text{C}_2\text{H}_2
ightarrow 4\ \text{mol }\text{CO}_2$, so ratio = $\frac{4}{2}=2$
$\text{Moles of }\text{CO}_2 = 0.826\ \text{mol} \times 2 = 1.652\ \text{mol}$

Step4: Theoretical mass of $\text{CO}_2$

Molar mass of $\text{CO}_2 = 12.01+(2\times16.00)=44.01\ \text{g/mol}$
$\text{Theoretical mass} = 1.652\ \text{mol} \times 44.01\ \text{g/mol} \approx 72.7\ \text{g}$

Step5: Calculate percent yield

$\text{Percent yield} = \frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100 = \frac{46.5\ \text{g}}{72.7\ \text{g}} \times 100 \approx 64.0\%$

(Question 7):

Step1: Apply percent yield formula

$\text{Percent yield} = \frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100$

Step2: Substitute given values

$\text{Percent yield} = \frac{3.12\ \text{g}}{4.51\ \text{g}} \times 100 \approx 69.2\%$

(Question 8):

Step1: Molar mass of $\text{LiOH}$

Molar mass of $\text{LiOH} = 6.94+16.00+1.008=23.948\ \text{g/mol}$

Step2: Moles of $\text{LiOH}$

$\text{Moles of }\text{LiOH} = \frac{45.3\ \text{g}}{23.948\ \text{g/mol}} \approx 1.892\ \text{mol}$

Step3: Mole ratio to $\text{LiHCO}_3$

From reaction: $1\ \text{mol }\text{LiOH}
ightarrow 1\ \text{mol }\text{LiHCO}_3$, so moles of $\text{LiHCO}_3 = 1.892\ \text{mol}$

Step4: Theoretical mass of $\text{LiHCO}_3$

Molar mass of $\text{LiHCO}_3 = 6.94+1.008+12.01+(3\times16.00)=67.958\ \text{g/mol}$
$\text{Theoretical mass} = 1.892\ \text{mol} \times 67.958\ \text{g/mol} \approx 128.6\ \text{g}$

(Question 9):

Step1: Rearrange percent yield formula

$\text{Theoretical yield} = \frac{\text{Actual yield}}{\text{Percent yield}} \times 100$

Step2: Substitute given values

$\text{Theoretical yield} = \frac{72.8\ \text{g}}{51.4} \times 100 \approx 141.6\ \text{g}$

Answer:

1. $\approx 64.0\%$
2. $\approx 69.2\%$
3. $\approx 129\ \text{g}$ (rounded to 3 significant figures)
4. $\approx 142\ \text{g}$ (rounded to 3 significant figures)
5. $\approx 92.7\%$