If \( 62.7 \mathrm{gCr}(\mathrm{s}) \) reacts with excess \( \mathrm{S}_{8}(\mathrm{~s}) \) and \( 144.7{\mathrm{~g} \text { of } \mathrm{Cr}_{2} \mathrm{~S}_{3}(\mathrm{~s}) \text { formed from an experimemt, }}_{\text {What is the percent yield for } \mathrm{Cr}_{2} \mathrm{~S}_{3}(\mathrm{~s}) \text { ? (answer to } 1 \text { decimal place) }}^{\text {MW } \mathrm{S}_{4}=256.8 \mathrm{~g} / \mathrm{mol}} \quad \mathrm{MW} \mathrm{Cr}_{2} \mathrm{~S}_{3}=200.19 \mathrm{~g} / \mathrm{mol} \) \( 16 \mathrm{Cr}(\mathrm{s})+3 \mathrm{~S}_{8}(\mathrm{~s}) \rightarrow 8 \mathrm{Cr}_{2} \mathrm{~S}_{3}(\mathrm{~s}) \) Your Answer:

To find the percent yield of \( \mathrm{Cr}_{2} \mathrm{S}_{3} \), we need to follow these steps: 1. **Determine the moles of \( \mathrm{Cr} \) used.** 2. **Use the stoichiometry of the reaction to find the theoretical yield of \( \mathrm{Cr}_{2} \mathrm{S}_{3} \).** 3. **Calculate the percent yield using the actual yield and theoretical yield.** ### Step 1: Calculate moles of \( \mathrm{Cr} \) Given: - Mass of \( \mathrm{Cr} = 62.7 \, \text{g} \) - Molar mass of \( \mathrm{Cr} = 51.996 \, \text{g/mol} \) (approximately) \[ \text{Moles of } \mathrm{Cr} = \frac{\text{mass}}{\text{molar mass}} = \frac{62.7 \, \text{g}}{51.996 \, \text{g/mol}} \] ### Step 2: Use stoichiometry to find theoretical yield of \( \mathrm{Cr}_{2} \mathrm{S}_{3} \) From the balanced equation: \[ 16 \mathrm{Cr} + 3 \mathrm{S}_{8} \rightarrow 8 \mathrm{Cr}_{2} \mathrm{S}_{3} \] This means: - 16 moles of \( \mathrm{Cr} \) produce 8 moles of \( \mathrm{Cr}_{2} \mathrm{S}_{3} \). Thus, the ratio of \( \mathrm{Cr} \) to \( \mathrm{Cr}_{2} \mathrm{S}_{3} \) is: \[ \frac{8 \, \text{moles of } \mathrm{Cr}_{2} \mathrm{S}_{3}}{16 \, \text{moles of } \mathrm{Cr}} = \frac{1}{2} \] So, for every 2 moles of \( \mathrm{Cr} \), we get 1 mole of \( \mathrm{Cr}_{2} \mathrm{S}_{3} \). ### Step 3: Calculate the theoretical yield of \( \mathrm{Cr}_{2} \mathrm{S}_{3} \) Now, we will calculate the moles of \( \mathrm{Cr} \) and then find the moles of \( \mathrm{Cr}_{2} \mathrm{S}_{3} \) produced. Let's calculate the moles of \( \mathrm{Cr} \) first: \[ \text{Moles of } \mathrm{Cr} = \frac{62.7}{51.996} \approx 1.204 \, \text{mol} \] Using the stoichiometry: \[ \text{Moles of } \mathrm{Cr}_{2} \mathrm{S}_{3} = \frac{1.204 \, \text{mol Cr}}{2} \approx 0.602 \, \text{mol} \] Now, we can find the theoretical mass of \( \mathrm{Cr}_{2} \mathrm{S}_{3} \): \[ \text{Mass of } \mathrm{Cr}_{2} \mathrm{S}_{3} = \text{moles} \times \text{molar mass} = 0.602 \, \text{mol} \times 200.19 \, \text{g/mol} \] ### Step 4: Calculate the actual yield and percent yield The actual yield of \( \mathrm{Cr}_{2} \mathrm{S}_{3} \) is given as \( 144.7 \, \text{g} \). Now, let's calculate the theoretical mass of \( \mathrm{Cr}_{2} \mathrm{S}_{3} \): \[ \text{Mass of } \mathrm{Cr}_{2} \mathrm{S}_{3} = 0.602 \times 200.19 \approx 120.5 \, \text{g} \] Now, we can calculate the percent yield: \[ \text{Percent Yield} = \left( \frac{\text{actual yield}}{\text{theoretical yield}} \right) \times 100 = \left( \frac{144.7 \, \text{g}}{120.5 \, \text{g}} \right) \times 100 \] Now, let's calculate the percent yield. \[ \text{Percent Yield} = \left( \frac{144.7}{120.5} \right) \times 100 \approx 120.0\% \] ### Final Answer The percent yield for \( \mathrm{Cr}_{2} \mathrm{S}_{3} \) is approximately \( 120.0\% \).