Silicon nitride (Si3N4) is made by combining Si and nitrogen gas at a high temperature. How much (in grams) of Si is needed to react with an excess of nitrogen gas to prepare 125 g of silicon nitride if the percent yield of the reaction is 95.0%?
The correct answer and explanation is :
To find the mass of silicon (Si) needed to prepare 125 g of silicon nitride (Si₃N₄) with a 95.0% yield, we follow these steps:
Step 1: Write the balanced chemical equation
3Si+2N2→Si3N43Si + 2N_2 \rightarrow Si_3N_4
This tells us that 3 moles of silicon react with 2 moles of nitrogen to form 1 mole of silicon nitride (Si₃N₄).
Step 2: Molar masses
- Si: 28.09 g/mol
- Si₃N₄: 3×28.09+4×14.01=84.27+56.04=140.31 g/mol3 \times 28.09 + 4 \times 14.01 = 84.27 + 56.04 = 140.31 \text{ g/mol}
Step 3: Adjust for percent yield
The actual yield is 125 g, but the reaction was only 95.0% efficient. So, we calculate the theoretical yield: Theoretical yield=125 g0.950=131.58 g\text{Theoretical yield} = \frac{125\text{ g}}{0.950} = 131.58\text{ g}
Step 4: Convert theoretical yield to moles of Si₃N₄
131.58 g140.31 g/mol=0.9375 mol of Si3N4\frac{131.58\text{ g}}{140.31\text{ g/mol}} = 0.9375\text{ mol of } Si_3N_4
Step 5: Use stoichiometry to find moles of Si needed
From the balanced equation:
3 mol Si : 1 mol Si₃N₄ 0.9375 mol Si3N4×3 mol Si1 mol Si3N4=2.8125 mol Si0.9375 \text{ mol Si}_3\text{N}_4 \times \frac{3 \text{ mol Si}}{1 \text{ mol Si}_3\text{N}_4} = 2.8125 \text{ mol Si}
Step 6: Convert moles of Si to grams
2.8125 mol Si×28.09 g/mol=79.01 g of Si2.8125 \text{ mol Si} \times 28.09 \text{ g/mol} = \boxed{79.01 \text{ g of Si}}
✅ Final Answer: 79.01 g of Si
💬 Explanation (300 words)
To determine how much silicon is needed to prepare 125 grams of silicon nitride (Si₃N₄) with a 95.0% reaction yield, we must work backward from the product to the reactant. The reaction between elemental silicon (Si) and nitrogen gas (N₂) produces Si₃N₄, and the balanced chemical equation shows a 3:1 molar ratio of Si to Si₃N₄.
First, we account for the percent yield. Since the reaction only gives 95.0% of the possible product, the actual yield (125 g) is less than the theoretical maximum. Dividing the actual yield by the percent yield (expressed as a decimal) gives us the theoretical yield, or how much product would form if the reaction went perfectly.
Next, we convert this theoretical mass of Si₃N₄ into moles using its molar mass (140.31 g/mol). With the number of moles of Si₃N₄ known, we use stoichiometry to determine how many moles of silicon are required. The equation shows that 3 moles of Si are needed to make 1 mole of Si₃N₄, so we multiply the moles of product by 3.
Finally, converting the required moles of silicon into grams using its molar mass (28.09 g/mol) gives us the final answer: 79.01 grams of silicon must be used when nitrogen is in excess, and the yield is 95.0%. This approach ensures we accurately adjust for both stoichiometry and reaction efficiency.
