using avogadros law

Question

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 9.4 m^3 of oxygen were consumed? Also, be sure your answer has a unit symbol, and is rounded to 2

 significant digits.

Answer 1

the underlying balanced equation here is

4NH3​(g) + 5O2​(g) → 6H2​O(g) + 4NO(g), meaning that the ratio of oxygen consumed to water produced is 5:6

you can use that ratio to find the volume of water produced from 9.4 m^3 of oxygen

9.4 m^3 x 6/5 = 11.28 m^3

rounding to 2 sig figs makes the final answer here 11 m^3

Answer 2

Write a correctly balanced equation for the reaction taking place:

4NH₃(g) + 5O₂(g) ==> 6H₂O(g) + 4NO(g) ... balanced equation

Since no temperature or pressure are given in the problem, we'll assume STP (1 atm, 273K)

We can simply use the stoichiometry of the balanced equation and from the liters of O₂, find the liters of H₂O that will form (assuming NH₃ is present in excess)

9.4m3 O₂ x 6 m³ H₂O / 5 m³ O₂ = 11.28 m³ H²O = 11 m³ H₂O (rounded to 2 sig. figs,)

Or...using more conventional units of liters (L), we have...

9.4m³ O₂ x 1000 L / 1 m³ = 9400 L O₂

9400 L O₂ x 6 L H₂O / 5 L O₂ = 11,280 L H₂O = 11,000 L H₂O (rounded to 2 sig. figs.)