Question

10 g of hydrogen and 64 g of oxygen were filled in a steel vessel and exploded. The amount of water formed in the reaction will be:
A.1 mole
B.2 moles
C.3 moles
D.4 moles

Answer 1

The question is based on the stoichiometry of the reactants taking part in the reactions and the Gay-Lussac’s law of combining volumes
Formula: ${\text{moles = }}\dfrac{{{\text{mass}}}}{{{\text{molar mass}}}}$

Complete step by step answer:
The reaction between hydrogen and oxygen to form water is as follows:
${\text{2}}{{\text{H}}_{\text{2}}}{\text{ + }}{{\text{O}}_{\text{2}}} \to {\text{2}}{{\text{H}}_{\text{2}}}{\text{O}}$
2 : 1 : 2
In the above reaction, 2 moles of hydrogen molecules reacts with one mole oxygen molecules to form 2 moles of water vapour, based on the Gay-Lussac’s law of combining volumes, which states that when two or more gases combine together to form products, then they do so in simple whole number ratios.
Expressing the above ratio in molecular weights, 4 grams of hydrogen (for 2 moles) react with 32 grams of oxygen (1 mole) to form 36 grams of water vapour (2 moles).
So for 10 grams of hydrogen:
Amount of oxygen required would be = $\dfrac{{32}}{4} \times 10$= 80 grams of oxygen
Since, only 64g of oxygen is present. Thus, oxygen is the limiting reactant.
On the other hand, 32 grams of oxygen form 36 grams of water
Therefore, 64 grams of oxygen = $\dfrac{{64}}{{32}} = 2$ moles of oxygen,
Since 1 mole of oxygen gives 2 moles of water vapour,
2 moles of oxygen should give 4 moles of water vapour.

Hence, the correct answer is option D.

Note:
Limiting reactant is that one which is completely consumed in the reaction and thus determines when the reaction is going to stop. It limits the reaction from continuing as there is none left to react with the in-excess reactant.