Question

# 20 ml of mixture of $$CO$$ and $${ C }_{ 2 }{ H }_{ 4 }$$ was exploded with 50 mL of $${ O }_{ 2 }$$. The volume after explosion was 45 mL of shaking with $$NaOH$$ solution only 15 mL $${ O }_{ 2 }$$ was left behind , what were the volume of $$CO$$ and $${ C }_{ 2 }{ H }_{ 4 }$$ in mixture?

A
CO=5mL, C2H2=15 mL
B
CO=10mL, C2H2=10 mL
C
CO=8mL, C2H2=12 mL
D
CO=15mL, C2H2=5 mL

Solution

## The correct option is B $$CO=10mL,\ { C }_{ 2 }{ H }_{ 2 }=10\ mL$$Solution:- Let the mixture contain $$x \; mL$$ of $${C}_{2}{H}_{4}$$ and $$y \; mL$$ of $$CO$$.$$\underset{x}{{C}_{2}{H}_{4}} + 3 {O}_{2} \longrightarrow 2 \underset{2x}{C{O}_{2}} + 2 {H}_{2}O$$$$\underset{y}{CO} + \cfrac{1}{2} {O}_{2} \longrightarrow \underset{y}{C{O}_{2}}$$$$NaOH$$ absorbs the $$C{O}_{2}$$ produced in the reaction.Total amount of $$C{O}_{2}$$ produced $$= 2x + y = \left( 45 - 15 \right)$$$$\Rightarrow 2x + y = 30 ..... \left( 1 \right)$$Given that:-$$x + y = 20 ..... \left( 2 \right)$$From $${eq}^{n} \left( 1 \right) \& \left( 2 \right)$$, we have$$20 + y = 30$$$$\Rightarrow y = 10 \; mL$$Substituting the value of $$y$$ in $${eq}^{n} \left( 2 \right)$$, we have$$x + 10 = 20$$$$\Rightarrow x = 10 \; mL$$Hence the volume of $$CO$$ and $${C}_{2}{H}_{4}$$ in the mixture is $$10 \; mL$$ and $$10 \; mL$$ respectively.Chemistry

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