68.93 g of water
The equation for the reaction is;
4NH3(g)+5O2(g)⟶4NO(g)+6H2O(g)
We are given;
Mass of ammonia = 51.0 g
Mass of Oxygen = 51.0 g
We are required to determine the maximum amount of water that can be produced.
Moles = Mass ÷ Molar mass
Moles of ammonia gas
Molar mass of ammonia = 17.031 g/mol
Moles = 51.0 g ÷ 17.031 g/mol
= 2.995 moles
= 3 moles
Moles of Oxygen gas
Molar mass of oxygen = 16.0 g/mol
Moles = 51.0 g ÷ 16.0 g/mol
= 3.1875 moles
But, from the reaction 4 moles of ammonia reacts with 5 moles of oxygen gas.
Therefore, Oxygen gas is the limiting reactant since the number of moles of oxygen gas are more compared to those of 3.1875 moles.
From the reaction, 5 moles of oxygen reacts to produce 6 moles of water.
Therefore, for 3.1875 moles of oxygen;
= 3.1875 moles × 6/5
= 3.825 moles of H₂O
Mass = number of moles × Molar mass
Molar mass of water = 18.02 g/mol
Therefore;
Moles of water = 3.825 moles × 18.02 g/mol
= 68.9265 g
= 68.93 g of water
Answer:
Mass of water = 34.56 g
Explanation:
Given data:
Mass of ammonia = 51.0 g
Mass of oxygen = 51.0 g
Mass of water produced = ?
Chemical equation:
4NH₃ + 5O₂ → 4NO + 6H₂O
Number of moles of oxygen:
Number of moles of oxygen = Mass /molar mass
Number of moles of oxygen = 51.0 g / 32 g/mol
Number of moles of oxygen = 1.6 mol
Number of moles of ammonia:
Number of moles of ammonia = Mass /molar mass
Number of moles of ammonia = 51.0 g / 17 g/mol
Number of moles of ammonia = 3 mol
Now we will compare the moles of water with ammonia and oxygen .
NH₃ : H₂O
4 : 6
3 : 6/4 ×3 = 4.5 mol
O₂ : H₂O
5 : 6
1.6 : 6/5 × 1.6= 1.92 mol
The number of moles of water produced by oxygen are less it will be limiting reactant.
Mass of water = moles × molar mass
Mass of water = 1.92 mol × 18 g/mol
Mass of water = 34.56 g
