When 2.00 g of Si reacts with 1.50 g of N_2 to form Si_3N_4 according to the equation 3 Si + 2 N_2 → Si_3N_4, which is the limiting reactant?

• A. N_2
• B. Neither is limiting
• C. Si
• D. Both are limiting

Answer: (C)

Verified step by step guidance

1. Write down the balanced chemical equation: \$3 \text{Si} + 2 \text{N}_2 \rightarrow \text{Si}_3\text{N}_4$.
2. Calculate the number of moles of each reactant using their molar masses: \(\text{moles of Si} = \frac{2.00\,\text{g}}{28.09\,\text{g/mol}}\) and \(\text{moles of } N_2 = \frac{1.50\,\text{g}}{28.02\,\text{g/mol}}\).
3. Use the mole ratio from the balanced equation to find the amount of product each reactant can produce. For Si, use \(\frac{1}{3}\) mole of \(\text{Si}_3\text{N}_4\) per mole of Si; for \(N_2\), use \(\frac{1}{2}\) mole of \(\text{Si}_3\text{N}_4\) per mole of \(N_2\).
4. Compare the amounts of \(\text{Si}_3\text{N}_4\) that can be formed from each reactant. The reactant that produces the lesser amount of product is the limiting reactant.
5. Conclude which reactant is limiting based on the comparison and explain that the limiting reactant determines the maximum amount of product formed.