Question

Consider the reaction n2 + 3 h2 → 2 nh3 . how much nh3 can be produced from the reaction of 74.2 g of n2 and 14.0 moles of h2? 1. 1.26 × 1025 molecules 2. 1.59 × 1024 molecules 3. 3.19 × 1024 molecules 4. 5.62 × 1024 molecules 5. 1.69 × 1025 molecules

Answer 1

The correct amount of NH3 that can be produced is 3.19 × 10^24 molecules, found by determining N2 as the limiting reactant and using stoichiometry.

Step-by-step explanation:

The question involves calculating the amount of ammonia (NH3) that can be produced from the reaction of 74.2 g of nitrogen (N2) and 14.0 moles of hydrogen (H2), given the balanced chemical equation N2(g) + 3H2(g) → 2NH3(g). This is a stoichiometry problem where we need to find the limiting reactant and use molar ratios to compute the number of molecules of NH3 produced.

• First, convert the mass of N2 to moles: 74.2 g N2 × (1 mol N2 / 28.02 g N2) = 2.65 moles N2.
• Determine the limiting reactant by comparing the mole ratio from the balanced equation to the moles available. For N2, it would require 2.65 moles × 3 = 7.95 moles of H2 but there are 14.0 moles of H2 available, so N2 is the limiting reactant.
• Using the mole ratio from the equation, calculate the moles of NH3 produced: 2.65 moles N2 × (2 moles NH3 / 1 mol N2) = 5.30 moles NH3.
• Convert moles of NH3 to molecules using Avogadro's number: 5.30 moles NH3 × (6.022 × 10^23 molecules / 1 mol) = 3.19 × 10^24 molecules NH3.

Therefore, option 3. 3.19 × 10^24 molecules is the correct answer.