Question

Ammonia is produced from the reaction of nitrogen and hydrogen according to the following balanced equation.

N2(g) + 3 H2(g) → 2 NH3(g)
A) What is the maximum mass (in g) of ammonia that can be produced from a mixture of 6.65 ✕ 10^2 g N2 and 1.10 ✕ 10^2 g H2?
B) What mass (in g) of which starting material would remain unreacted?"

Answer 1

The maximum mass of ammonia produced from the given reactants is 623.39 g, with hydrogen as the limiting reagent. After the reaction, 151.76 g of nitrogen, the excess reactant, would remain unreacted.

Step-by-step explanation:

Calculating Mass of Ammonia Produced and Excess Reactants

To calculate the maximum mass of ammonia that can be produced from a mixture of nitrogen (N2) and hydrogen (H2), we need to determine the limiting reagent and use stoichiometry based on the balanced chemical equation: N2(g) + 3 H2(g) → 2 NH3(g).

1. Convert the masses of N2 and H2 to moles using their molar masses (N2: 28 g/mol, H2: 2 g/mol).
2. For 6.65 × 102 g N2: moles of N2 = 6.65 × 102 g / 28 g/mol = 23.75 mol of N2.
3. For 1.10 × 102 g H2: moles of H2 = 1.10 × 102 g / 2 g/mol = 55 mol of H2.
4. Use the molar ratio from the balanced equation to determine the maximum moles of NH3 produced from each reactant.
5. Moles of NH3 from N2 = 23.75 mol N2 × (2 mol NH3 / 1 mol N2) = 47.5 mol NH3.
6. Moles of NH3 from H2 = 55 mol H2 × (2 mol NH3 / 3 mol H2) = 36.67 mol NH3.
7. The smallest mole value actually determines the maximum amount of NH3 that can be produced, which in this case is 36.67 mol NH3, indicating H2 is the limiting reagent.
8. Convert the moles of NH3 back to grams using its molar mass (NH3: 17 g/mol).
9. Maximum mass of NH3 = 36.67 mol × 17 g/mol = 623.39 g.
10. To find out how much of the excess reagent, N2, remains, subtract the amount that reacted (based on the limiting reagent H2).
11. Moles of N2 that reacted = 55 mol H2 / (3 mol H2 / 1 mol N2) = 18.33 mol N2.
12. The original amount of N2 was 23.75 mol, therefore, remaining moles of N2 = 23.75 mol - 18.33 mol = 5.42 mol.
13. Convert the remaining moles of N2 to grams: 5.42 mol × 28 g/mol = 151.76 g.

Thus, the maximum mass of ammonia that can be produced is 623.39 g, and 151.76 g of nitrogen would remain unreacted.