`10 g` of` S` reacts with excess of `O_2` to form` 15 g` of` SO_2`. The` %` yield of the reaction is

Options

A. 0.25
B. 0.5
C. 0.75
D. 1

Answer

To find the percentage yield of the reaction where 10 g of sulfur reacts with excess oxygen to form 15 g of sulfur dioxide (SO₂), follow these steps:

### Step 1: Write the balanced chemical equation
The reaction between sulfur (S) and oxygen (O₂) to form sulfur dioxide (SO₂) can be represented as:
\[ S + O_2 \rightarrow SO_2 \]

### Step 2: Calculate the moles of sulfur
To find the moles of sulfur, we use the formula:
\[ \text{Moles} = \frac{\text{mass}}{\text{molar mass}} \]
The molar mass of sulfur (S) is 32 g/mol. Therefore, the moles of sulfur in 10 g is:
\[ \text{Moles of S} = \frac{10 \, \text{g}}{32 \, \text{g/mol}} = 0.3125 \, \text{mol} \]

### Step 3: Determine the theoretical yield of SO₂
From the balanced equation, 1 mole of sulfur produces 1 mole of sulfur dioxide. Therefore, the moles of SO₂ produced from 0.3125 moles of sulfur will also be 0.3125 moles.

Now, calculate the mass of SO₂ produced using its molar mass (64 g/mol):
\[ \text{Mass of SO₂} = \text{moles} \times \text{molar mass} = 0.3125 \, \text{mol} \times 64 \, \text{g/mol} = 20 \, \text{g} \]

### Step 4: Calculate the percentage yield
The percentage yield is calculated using the formula:
\[ \text{Percentage Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100 \]
Here, the actual yield is given as 15 g, and the theoretical yield we calculated is 20 g:
\[ \text{Percentage Yield} = \left( \frac{15 \, \text{g}}{20 \, \text{g}} \right) \times 100 = 75\% \]

### Final Answer
The percentage yield of the reaction is **75%**.

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