Question

What is the limiting reactant and how much NO₂ is produced when 43.9 g of NH₃ reacts with 258 g of O2?

(molar masses: NH₃ = 17.03 g/mol, O₂= 32.0 g/mol, NO₂ = 46.01 g/mol)
4 NH₃(g) + 7 O₂(g) → 4NO₂(g) + 6 H₂O(g)
O NH₃ is the limiting reactant; 119 g NO₂ is produced.
O O₂ is the limiting reactant; 258 g NO₂ is produced.
O O₂ is the limiting reactant; 212 g NO₂ is produced.
O NH₃ is the limiting reactant; 212 g NO₂ is produced.
O O₂ is the limiting reactant; 119 g NO₂ is produced.

Answer 1

NH₃ is the limiting reactant in the reaction, with approximately 2.58 moles available option (a). Calculations based on stoichiometry show that NH₃ is consumed completely while O₂ is in excess. Therefore, 119 g of NO₂ is produced.

Step-by-step explanation:

To determine the limiting reactant in the reaction of NH₃ with O₂, we must calculate the moles of each reactant and then compare the stoichiometric ratio dictated by the balanced equation.

The balanced chemical equation is: 4NH₃(g) + 7O₂(g) → 4NO₂(g) + 6H₂O(g).

To find the moles of NH₃ and O₂, we use their given masses and molar masses:

• Moles of NH₃ = 43.9 g / 17.03 g/mol = approximately 2.58 moles.
• Moles of O₂ = 258 g / 32.0 g/mol = approximately 8.06 moles.

Now, we compare the moles of each reactant to the stoichiometric coefficients in the balanced equation. For every 4 moles of NH₃, 7 moles of O₂ are required. Thus:

• Required moles of O₂ for 2.58 moles of NH₃ = (2.58 moles NH₃) × (7 moles O₂ / 4 moles NH₃) ≈ 4.51 moles of O₂.

Since we have 8.06 moles of O₂ available, which is more than the required 4.51 moles, NH₃ is the limiting reactant.

Now, we can calculate the amount of NO₂ produced from the limiting reactant:

• Moles of NO₂ produced = (2.58 moles NH₃) × (4 moles NO₂ / 4 moles NH₃) = 2.58 moles of NO₂.
• Mass of NO₂ produced = 2.58 moles × 46.01 g/mol = 118.75 g, which we can round to 119 g.

NH₃ is the limiting reactant; 119 g NO₂ is produced.