Antimony reacts with sulphur according to the equation
`2Sb(s) + 3S(s) to Sb_(2)S_(3)(s)`
The molar mass of `Sb_(2)S_(3)` is `340 g mol^(-)1`
What is the percentage yield for a reaction in which 1.40 g of `Sb_(2)S_(3)` is obtained from 1.73 g of antimony and a slight excess of sulphur?

Options

A. 80.9
B. 58
C. 40.5
D. 29.0%

Explanation

To calculate the percentage yield of the reaction where antimony reacts with sulfur to form antimony trisulfide (Sb₂S₃), we will follow these steps:

### Step 1: Write down the balanced chemical equation
The balanced equation for the reaction is:
\[ 2 \text{Sb}(s) + 3 \text{S}(s) \rightarrow \text{Sb}_2\text{S}_3(s) \]

### Step 2: Calculate the molar mass of antimony (Sb)
The molar mass of antimony (Sb) is approximately 122 g/mol.

### Step 3: Calculate the moles of antimony used
Given that we have 1.73 g of antimony, we can calculate the number of moles of antimony:
\[ \text{Moles of Sb} = \frac{\text{mass of Sb}}{\text{molar mass of Sb}} = \frac{1.73 \, \text{g}}{122 \, \text{g/mol}} \approx 0.0142 \, \text{mol} \]

### Step 4: Use stoichiometry to find the theoretical yield of Sb₂S₃
From the balanced equation, 2 moles of Sb produce 1 mole of Sb₂S₃. Therefore, we can find the moles of Sb₂S₃ produced from the moles of Sb:
\[ \text{Moles of Sb}_2\text{S}_3 = \frac{0.0142 \, \text{mol Sb}}{2} \approx 0.0071 \, \text{mol Sb}_2\text{S}_3 \]

### Step 5: Calculate the theoretical mass of Sb₂S₃
The molar mass of Sb₂S₃ is given as 340 g/mol. Thus, the theoretical mass of Sb₂S₃ can be calculated as:
\[ \text{Mass of Sb}_2\text{S}_3 = \text{moles of Sb}_2\text{S}_3 \times \text{molar mass of Sb}_2\text{S}_3 = 0.0071 \, \text{mol} \times 340 \, \text{g/mol} \approx 2.41 \, \text{g} \]

### Step 6: Calculate the percentage yield
The experimental yield of Sb₂S₃ is given as 1.40 g. The percentage yield can be calculated using the formula:
\[ \text{Percentage Yield} = \left( \frac{\text{Experimental Yield}}{\text{Theoretical Yield}} \right) \times 100 \]
Substituting the values:
\[ \text{Percentage Yield} = \left( \frac{1.40 \, \text{g}}{2.41 \, \text{g}} \right) \times 100 \approx 58.0\% \]

### Final Answer
The percentage yield of the reaction is approximately **58%**.

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