How many grams of NH3 can be produced from 2.55 mol of N2 and excess H2? How many grams of H2 are needed to produce 10.38 g of NH3? How many molecules (not moles) of NH3 are produced from 7.87×10^-4 g of H2?
The Correct Answer and Explanation is:
To answer the questions, we need to use the balanced chemical equation for the synthesis of ammonia (NH₃):
N₂ + 3H₂ → 2NH₃
1. How many grams of NH₃ can be produced from 2.55 mol of N₂ and excess H₂?
From the balanced equation:
- 1 mol of N₂ produces 2 mol of NH₃
- So, 2.55 mol of N₂ will produce:
2.55 mol N₂ × (2 mol NH₃ / 1 mol N₂) = 5.10 mol NH₃
Now convert moles of NH₃ to grams:
- Molar mass of NH₃ = 14.01 (N) + 3 × 1.008 (H) = 17.034 g/mol
- Mass = 5.10 mol × 17.034 g/mol = 86.87 g NH₃
2. How many grams of H₂ are needed to produce 10.38 g of NH₃?
First, convert grams of NH₃ to moles:
- 10.38 g ÷ 17.034 g/mol = 0.6096 mol NH₃
From the balanced equation:
- 2 mol NH₃ are produced from 3 mol H₂
- 0.6096 mol NH₃ × (3 mol H₂ / 2 mol NH₃) = 0.9144 mol H₂
Convert moles of H₂ to grams:
- Molar mass of H₂ = 2.016 g/mol
- 0.9144 mol × 2.016 g/mol = 1.844 g H₂
3. How many molecules of NH₃ are produced from 7.87×10⁻⁴ g of H₂?
Convert grams of H₂ to moles:
- 7.87×10⁻⁴ g ÷ 2.016 g/mol = 3.905×10⁻⁴ mol H₂
From the equation:
- 3 mol H₂ produce 2 mol NH₃
- 3.905×10⁻⁴ mol H₂ × (2 mol NH₃ / 3 mol H₂) = 2.603×10⁻⁴ mol NH₃
Convert moles to molecules:
- Use Avogadro’s number: 6.022×10²³ molecules/mol
- 2.603×10⁻⁴ mol × 6.022×10²³ = 1.568×10²⁰ molecules of NH₃
Explanation
To solve chemical problems like these, it is essential to begin with a balanced chemical equation. The synthesis of ammonia, a common reaction in industrial chemistry, follows this formula: N₂ + 3H₂ → 2NH₃. This shows that one molecule of nitrogen reacts with three molecules of hydrogen to form two molecules of ammonia. The stoichiometric coefficients (1, 3, and 2) help determine the mole ratios needed for calculations.
In the first part, we are given 2.55 mol of nitrogen gas and asked to find how much ammonia is produced, assuming hydrogen is in excess. Since one mole of nitrogen produces two moles of ammonia, multiplying 2.55 mol by 2 gives 5.10 mol NH₃. Using ammonia’s molar mass, we convert moles to grams.
The second question starts with a mass of ammonia, and we work backward. First, we convert that mass to moles using the molar mass of ammonia. Then, we use the mole ratio to find how many moles of hydrogen gas are required. Finally, converting those moles to grams gives the mass of hydrogen needed.
In the final question, we are given a tiny mass of hydrogen gas and asked how many ammonia molecules it can produce. After converting grams of hydrogen to moles, we again apply the mole ratio to find the number of moles of ammonia produced. We then multiply by Avogadro’s number to get the exact number of ammonia molecules.
These problems demonstrate the importance of understanding mole ratios, molar masses, and the conversion between moles and molecules in chemistry.
