Final answer:
The theoretical yield of ammonia when 59.8 g N2 reacts with 5.2 g H2 is 29.1 grams, calculated based on stoichiometry and the balanced chemical equation.
Step-by-step explanation:
The theoretical yield of ammonia (NH3) is calculated using stoichiometry based on the balanced chemical equation N2 + 3 H2 → 2 NH3. To find the theoretical yield of ammonia when 59.8 g of nitrogen (N2) react with 5.2 g of hydrogen (H2), first, we need to determine the moles of each reactant. Using the molar mass of nitrogen (28.02 g/mol) and hydrogen (2.02 g/mol), we calculate:
- Moles of N2 = 59.8 g / 28.02 g/mol = 2.14 mol
- Moles of H2 = 5.2 g / 2.02 g/mol = 2.57 mol
Next, we determine the limiting reactant. Since 1 mol of N2 reacts with 3 mol of H2, for 2.14 mol of N2, we would need 3 × 2.14 mol = 6.42 mol of H2. However, we only have 2.57 mol of H2, hence hydrogen is the limiting reactant.
According to the balanced equation, 3 moles of H2 produce 2 moles of NH3. Thus, 2.57 mol H2 would produce (2.57 mol H2 × (2 mol NH3 / 3 mol H2)) = 1.71 mol NH3. Finally, by multiplying this by the molar mass of ammonia (17.03 g/mol), we find the theoretical yield in grams:
Theoretical yield of NH3 = 1.71 mol × 17.03 g/mol = 29.1 g