4NH3(g) + 7O2(g) → 4NO2(g) + 6H2O(g), How many grams of ammonia will be required to produce 12.0 g of water?

Answer 1

First, write the mole ratio of ammonia to water: 4 mol NH3/ 6 mol H2O

Convert the grams H2O to moles: 1 mol H2O / 18.0 g H2O x 12.0 g H2O

Write the conversion factor from moles NH3 to grams NH3: 17 g NH3/ 1 mol NH3

Multiply all three conversion factors: (cancel similar units)

1 mol H2O / 18.0 g H2O x 12.0 g H2O x 4 mol NH3./ 6 mol H2O x 17 g NH3/ 1 mol NH3 = 7.56 g NH3 All units cancel except for g NH3.

Answer 2

Hello, Addy,

The equation is balanced, so we know that it takes 4 moles of NH₃ to make 6 moles of H₂O, a molar ratio of 4/6 or 2/3. [Or, we could also say 6 moles of H₂O comes from 4 moles of NH₃, a ratio of 6/4, or 1.5].

12 grams of water are produced. Convert grams into moles by dividing by water's molar mass of 18 g/mole. We have 2/3 or 0.6667 moles of H₂O. Multiply this by the molar ratio of 2/3, since we only need 2 moles of NH₃ to make 3 moles of H₂O. That leads to 0.44447 moles of NH₃ required to produce 0.6667 moles (12 grams) of water. Multiply by the molar mass of ammonia to give 7.6 grams of NH₃ (2 sig figs).

Bob