Calculate the percentage yield of the reaction if 64 g of`NaBH_(4)` with iodine produced 15.0 g of`BI_(3)`.
`NaBH_(4) + 4I_(2) to BI_(3) + NaI + 4HI`
(At. mass, Na = 23, B = 10.8,1 = 127)

Explanation

To calculate the percentage yield of the reaction between sodium borohydride (NaBH₄) and iodine (I₂) producing bismuth triiodide (BiI₃), we will follow these steps:

### Step 1: Write the balanced chemical equation
The balanced equation for the reaction is:
\[ \text{NaBH}_4 + 4 \text{I}_2 \rightarrow \text{BiI}_3 + \text{NaI} + 4 \text{HI} \]

### Step 2: Calculate the molar masses
- Molar mass of Na: 23 g/mol
- Molar mass of B: 10.8 g/mol
- Molar mass of I: 127 g/mol

Calculating the molar mass of NaBH₄:
\[ \text{Molar mass of NaBH}_4 = 23 + 10.8 + (4 \times 1) = 23 + 10.8 + 4 = 37.8 \text{ g/mol} \]

Calculating the molar mass of BiI₃:
\[ \text{Molar mass of BiI}_3 = 10.8 + (3 \times 127) = 10.8 + 381 = 391.8 \text{ g/mol} \]

### Step 3: Determine the theoretical yield of BiI₃
From the balanced equation, 1 mole of NaBH₄ produces 1 mole of BiI₃.

If 37.8 g of NaBH₄ produces 391.8 g of BiI₃, we can find out how much BiI₃ can be produced from 64 g of NaBH₄ using a unitary method:
\[ \text{Theoretical yield of BiI}_3 = \left( \frac{64 \text{ g NaBH}_4}{37.8 \text{ g NaBH}_4} \right) \times 391.8 \text{ g BiI}_3 \]
Calculating this:
\[ \text{Theoretical yield of BiI}_3 = \left( \frac{64}{37.8} \right) \times 391.8 \approx 663.3651 \text{ g} \]

### Step 4: Calculate the percentage yield
The percentage yield is calculated using the formula:
\[ \text{Percentage yield} = \left( \frac{\text{Actual yield}}{\text{Theoretical yield}} \right) \times 100 \]
Given that the actual yield of BiI₃ is 15.0 g:
\[ \text{Percentage yield} = \left( \frac{15.0 \text{ g}}{663.3651 \text{ g}} \right) \times 100 \]
Calculating this:
\[ \text{Percentage yield} \approx \left( 0.0226 \right) \times 100 \approx 2.26\% \]

### Final Answer
The percentage yield of the reaction is approximately **2.26%**.

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