Question

4 mol NH3 + 5 mol oxygen gas -> 4 mol nitrogen monoxide+ Mol water if 172.68 grams of nitrogen monoxide are produced, how many grams of water will also be produced

Answer 1

Molar mass Nitrogen monoxide = 30.01

Moles of Nitrogen monoxide $= \frac{172.68}{30.01}=5.75 moles$

4NH₃+5O₂---->4NO+6H₂O

Moles of H2O $=\frac{6×5.75}{4}=8.625$ moles

Molar mass of water is 18.02

Mass of H2O = $8.625× 18.02=155.42 g$