Question

If $8$ mL of uncombined $O_2$ remain after exploding $O_2$ with $4$ mL of hydrogen, the volume in mL of $O_2$ originally was :

• A. $12$
• B. $2$
• C. $10$
• D. $4$

Answer 1

The correct option is C $10$

The balanced chemical reaction is $\frac{1}{2}{O}_2\left(g\right)+H_2\left(g\right)⟶H_{2}O$.
Initially, say $V$ mL of oxygen was present.
The volume of hydrogen present initially was $4$ mL.
After reaction, $V_2$ mL of oxygen and $0$ mol of hydrogen are present.
However, $8$ mL of oxygen remained after the reaction.
Thus, $V_2=8$ mL or, $V=10$ mL.