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10 g of hydrogen and 64 g of oxygen were filled in a steel vessel and exploded. Amount of water produced in this reaction will be -

Option: 1

1 mol


Option: 2

2 mol


Option: 3

3 mol


Option: 4

4 mol


Answers (1)

best_answer

The reaction will be-

\mathrm{2 H_{2}(g)+O_{2}(g)\rightarrow 2 H_{2} O(l) }

Two moles of Hand one mole of Owill produce two moles of water.

\mathrm{\text{No. of moles in 10g of } H_{2} = \frac{\text{given mass of } H_{2}}{\text{molar mass of }H_{2}} =\frac{10}{2} = 5 \text{ moles}}

\mathrm{\text{No. of moles in 64g of } O_{2} = \frac{\text{given mass of } O_{2}}{\text{molar mass of }O_{2}} =\frac{64}{32} = 2 \text{ moles}}

5 Moles of H2 can produce 5 moles of water while 2 moles of O2 can produce 4 moles of water. 

Thus, O2 is the limiting reactant and 4 moles H2O will be produced.

Therefore, the correct option is (4).

Posted by

Rakesh

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