Answer:
Mass of excess reactant left = 179.6 g
Limiting reactant = nitrogen
Mass of ammonia formed = 200.6 g
Explanation:
Given data:
Mass of nitrogen = 165.0 g
Mass of hydrogen = 215.0 g
Limiting reactant = ?
Mass of ammonia formed = ?
Mass of excess reactant left = ?
Solution:
Chemical equation:
N₂ + 3H₂ → 2NH₃
Number of moles of nitrogen:
Number of moles = mass/molar mass
Number of moles = 165.0 g/ 28 g/mol
Number of moles = 5.9 mol
Number of moles of hydrogen:
Number of moles = mass/molar mass
Number of moles = 215.0 g/ 2 g/mol
Number of moles = 107.5 mol
Now we will compare the moles of ammonia with both reactant.
H₂ : NH₃
3 : 2
107.5 : 2/3×107.5 = 71.7 mol
N₂ : NH₃
1 : 2
5.9 : 2/1×5.9 = 11.8 mol
Less number of moles of ammonia are formed by the nitrogen it will act as limiting reactant.
Mass of ammonia formed:
Mass = number of moles × molar mass
Mass = 11.8 mol × 17 g/mol
Mass = 200.6 g
Mass of hydrogen left:
We will compare the moles of hydrogen and nitrogen.
N₂ : H₂
1 : 3
5.9 : 3/1×5.9 = 17.7 mol
Out of 107.5 moles 17.7 moles of hydrogen react with nitrogen.
Number of moles left unreacted = 107.5 - 17.7 mol = 89.8 mol
Mass of hydrogen left:
Mass = number of moles × molar mass
Mass = 89.8 mol × 2 g/mol
Mass = 179.6 g
