20 mL of a mixture of C2H2 and C2H4 required 55 mL of O2 for complete combustion (all volumes at the same T and P). The volume of C2H2 in the mixture was

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$20 mL$ of a mixture of $C X_{2} H X_{2}$ and $C X_{2} H X_{4}$ required $55 mL$ of $O X_{2}$ for complete combustion (all volumes at the same $T$ and $P$ ). The volume of $C X_{2} H X_{2}$ in the mixture was

C X_{2} H X_{2}

needs

2.5

volumes of

O X_{2}

C X_{2} H X_{4}

needs

3

. Let

C X_{2} H X_{2} = a

2.5 a + 3 (20 - a) = 55

, so

60 - 0.5 a = 55

a = 10 mL