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
