Ammonia gas and oxygen gas react to form water vapor and nitrogen monoxide gas. What volume of water would be produced by this reaction if 6.3L of ammonia were consumed? Be sure your answer has the correct number of significant digits.

Water volume produced is 7.3 mL

Explanation:

This the reaction:

4NH₃ (g) + 5O₂ (g) → 4NO (g) + 6H₂O (g)

We have to work with density, to solve this question

Density of ammonia = Ammonia mass / Ammonia volume

0,00073 g/mL = Ammonia mass / 6300 mL

(Notice, we had to convert 6.3L to mL)

0.00073 g/mL. 6300 mL = Ammonia mass → 4.599 g

Mass / Molar mass = Mol

4.599 g / 17g/m = 0.270 mole

Ratio is 4:6

4 mole of ammonia produce 6 mole of water

0.270 mole produce (0.270.6) / 4 = 0.405 mol

Molar mass. mole = mass

18 g/m. 0.405 m = 7.30 g

Water density = 1 g/mL

Water density = Water mass / Water Volume

1g/mL = 7.30 g / Water volume

Water volume = 7.3 mL