If 30 mL of `H_(2)` and 20 mL of `O_(2)` reacts to form water, what is left at the end of the reaction?

Options

A. 10 mL of `H_(2)`
B. 5 mL of `H_(2)`
C. 10 mL of `O_(2)`
D. 5 mL of `O_(2)`

Answer

D

Explanation

`underset(2vol)(2H_(2))+underset(1vol)(O_(2))to2H_(2)O`
2 mole of `H_(2)` reacts with 1 vol of `O_(2)` to give water. Therefore, 30 mL of `H_(2)` will reacts with 15 mL of `O_(2)` to give water. Here, `H_(2)` is the limiting reactant.
Volume of `O_(2)` left unreacted = 20-15=5 mL