1.

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?

Answer»

10 mL of `H_(2)`
5 mL of `H_(2)`
10 mL of `O_(2)`
5 mL of `O_(2)`

Solution :`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


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