Question

A sample of gas contains 0.1900 mol of H2(g) and 9.500×10-2 mol of O2(g) and occupies a volume of 18.0 L. The following reaction takes place: 2H2(g) + O2(g)=2H2O(g) Calculate the volume of the sample after the reaction takes place, assuming that the temperature and the pressure remain constant.

Answer 1

Balanced reaction is

2H₂ + O₂   $\rightarrow$ 2H₂O

According to balanced reaction 2 mole H₂ react with 1 mole O₂ molar ratio between H₂ to O₂ is 2:1 thus to react with 0.1900 mole H₂ required mole of O₂ = 0.1900 / 2 = 9.5 X 10^-2 mole so reactant given in proportion.

2 mole H₂ produce 2 mole H₂O molar ratio between H₂ to H₂O is 1:1 therefore 0.1900 mole H₂ produce 0.1900 mole H₂O

mole of product H₂O produced = 0.1900 mole

Total mole of reactant = (mole of H₂) + (mole of O₂) = (0.1900) + (9.500 X 10^-2) = 0.285 mole

Total mole of reactant = 0.285 mole

We know the Avogadro's law at constant temperature and pressure volume is directly proportional to no. of moles

V₁/n₁ = V₂/n₂

Where,

V₁ = initial volume = 18 L

n₁ = initial mole = 0.285 mole

V₂ = final volume =?

n₂ = final mole = 0.1900 moles

We can write above formula as

V₂ = V₁n₂/ n₁

Substitute the value

V₂ = 18 L X 0.1900 mol / 0.285 mol = 12 L

After reaction take place volume of gas = 12 L