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Question: Find the mass of h2 required and NH3 produce form 14 gm of n2...

Find the mass of h2 required and NH3 produce form 14 gm of n2

Answer

The mass of H₂ required is 3 g. The mass of NH₃ produced is 17 g.

Explanation

Solution

To solve this problem, we need to use the principles of stoichiometry based on the balanced chemical equation for the synthesis of ammonia (Haber process).

1. Write the balanced chemical equation: The reaction between nitrogen (N₂) and hydrogen (H₂) to form ammonia (NH₃) is: N2(g)+3H2(g)→2NH3(g)\text{N}_2\text{(g)} + 3\text{H}_2\text{(g)} \rightarrow 2\text{NH}_3\text{(g)}

2. Calculate the molar masses of the substances involved:

  • Molar mass of N₂ = 2 × 14 g/mol = 28 g/mol
  • Molar mass of H₂ = 2 × 1 g/mol = 2 g/mol
  • Molar mass of NH₃ = 14 + (3 × 1) g/mol = 17 g/mol

3. Convert the given mass of N₂ to moles: Given mass of N₂ = 14 g Moles of N2=MassMolar mass=14 g28 g/mol=0.5 mol\text{Moles of N}_2 = \frac{\text{Mass}}{\text{Molar mass}} = \frac{14 \text{ g}}{28 \text{ g/mol}} = 0.5 \text{ mol}

4. Use mole ratios from the balanced equation to find moles of H₂ required and NH₃ produced: From the balanced equation:

  • 1 mole of N₂ reacts with 3 moles of H₂.
  • 1 mole of N₂ produces 2 moles of NH₃.

Therefore:

  • Moles of H₂ required = 0.5 mol N2×3 mol H21 mol N2=1.5 mol H20.5 \text{ mol N}_2 \times \frac{3 \text{ mol H}_2}{1 \text{ mol N}_2} = 1.5 \text{ mol H}_2
  • Moles of NH₃ produced = 0.5 mol N2×2 mol NH31 mol N2=1.0 mol NH30.5 \text{ mol N}_2 \times \frac{2 \text{ mol NH}_3}{1 \text{ mol N}_2} = 1.0 \text{ mol NH}_3

5. Convert moles of H₂ and NH₃ back to mass:

  • Mass of H₂ required: Mass of H2=Moles of H2×Molar mass of H2=1.5 mol×2 g/mol=3 g\text{Mass of H}_2 = \text{Moles of H}_2 \times \text{Molar mass of H}_2 = 1.5 \text{ mol} \times 2 \text{ g/mol} = 3 \text{ g}
  • Mass of NH₃ produced: Mass of NH3=Moles of NH3×Molar mass of NH3=1.0 mol×17 g/mol=17 g\text{Mass of NH}_3 = \text{Moles of NH}_3 \times \text{Molar mass of NH}_3 = 1.0 \text{ mol} \times 17 \text{ g/mol} = 17 \text{ g}

Explanation of the solution:

  1. Write the balanced chemical equation: N2+3H2→2NH3\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3.
  2. Calculate moles of given reactant (N2\text{N}_2): 14 g/28 g/mol=0.5 mol14 \text{ g} / 28 \text{ g/mol} = 0.5 \text{ mol}.
  3. Use stoichiometric ratios from the balanced equation to find moles of required reactant (H2\text{H}_2) and produced product (NH3\text{NH}_3).
    • Moles of H2=0.5 mol N2×(3 mol H2/1 mol N2)=1.5 mol H2\text{H}_2 = 0.5 \text{ mol N}_2 \times (3 \text{ mol H}_2 / 1 \text{ mol N}_2) = 1.5 \text{ mol H}_2.
    • Moles of NH3=0.5 mol N2×(2 mol NH3/1 mol N2)=1.0 mol NH3\text{NH}_3 = 0.5 \text{ mol N}_2 \times (2 \text{ mol NH}_3 / 1 \text{ mol N}_2) = 1.0 \text{ mol NH}_3.
  4. Convert moles back to mass:
    • Mass of H2=1.5 mol×2 g/mol=3 g\text{H}_2 = 1.5 \text{ mol} \times 2 \text{ g/mol} = 3 \text{ g}.
    • Mass of NH3=1.0 mol×17 g/mol=17 g\text{NH}_3 = 1.0 \text{ mol} \times 17 \text{ g/mol} = 17 \text{ g}.