Question

A sample of gas contains 0.1700 mol of NH3(g) and 0.2125 mol of O2(g) and occupies a volume of 17.8 L. The following reaction takes place: 4NH3(g) + 5O2(g)4NO(g) + 6H2O(g) Calculate the volume of the sample after the reaction takes place, assuming that the temperature and the pressure remain constant.

Answer 1

Answer:

The volume of the sample after the reaction takes place is 19.78 L.

Explanation:

The given variables are;

Number of moles of NH₃(g) = 0.1700 mol

Number of moles of O₂(g) = 0.2125 mol

Volume occupied by the mixture = 17.8 L

The reaction

4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g)

Then takes place

That is 4 moles of NH₃(g) reacts with 5 moles of O₂(g) to produce 4 moles of NO(g) and 6 moles of H₂O(g).

Since there are less number of moles of NH₃(g) (= 0.1700 mol) in the mixture, we factor the above equation by the number of moles of NH₃(g)  present.

That is,

1 moles of NH₃(g) reacts with 5/4 moles of O₂(g) to produce 1 moles of NO(g) and 3/2 moles of H₂O(g).

Therefore,

0.1700 mol of NH₃(g) reacts with 5/4×0.1700  moles of O₂(g) to produce 0.1700  moles of NO(g) and 3/2×0.1700  moles of H₂O(g).

Which gives

0.1700 mol of NH₃(g) reacts with 0.2125  moles of O₂(g) to produce 0.1700  moles of NO(g) and 0.255  moles of H₂O(g).

Therefore, all of the NH₃(g) and O₂(g)  are consumed in the reaction and the present gases in sample then becomes

0.1700  moles of NO(g) and 0.255  moles of H₂O(g).

Total number of moles of reactant = 0.17 + 0.2125 = 0.3825

Total number of moles of product formed = 0.17 + 0.255 = 0.425

However, Avogadro's law states that equal volume of all gases at the same temperature and pressure contains equal number of molecules.

That is volume occupied by  0.3825 moles of gas = 17.8 L

Therefore the volume occupied by  0.425 moles of gas = 17.8×0.425/0.3825 L = 19.78 L