Question

A volume of $100ml$ of water gas containing some $C{O}_2$ was mixed with $100ml$ of oxygen and mixture exploded. The volume after an explosion was $100ml$. On introducing $NaOH$, the volume was reduced to $52.5ml$. If the volume ratio of $CO$, $H_2$ and $C{O}_2$ in the original sample is $ab:cd:2$, calculate the value of ${}^{\prime }abc{d}^{\prime }$.

Answer 1

Let,

Vol. of $CO=xml$
Vol. of $H_2=yml$ &

Vol. of $C{O}_2=zml$

Given that,

$x+y+z=100ml$

Now,

$CO+\frac{1}{2}{O}_2⟶C{O}_2$

$x$ $\frac{1}{2}x$ $xml$

Again,

$H_2+\frac{1}{2}{O}_2⟶H_{2}O$

$y$ $\frac{1}{2}y$ Negligible volume

Total $O_2$ used $=\frac{x}{2}+\frac{y}{2}=\frac{x+y}{2}$

Total $C{O}_2=x+z$

Now, before and after reaction with oxygen,

$CO+H_2+C{O}_2+(\frac{x}{2}+\frac{y}{2}{O}_2)=C{O}_2+H_{2}O+$ unreacted $O_2$
$x$ $y$ $z$ $100ml$ $(x+z)$ Neg. $52.5ml=100-\frac{x+y}{2}$

After explosion total volume $=100ml$

$(x+z)+100-\left(\frac{x+y}{2}\right)=100$

$x+z=\frac{x+y}{2}$

$2x+2z=x+y$

$x=y-2z$ -------- Eqn. $1$

After the explosion, volume of $C{O}_2=$ Total volume $-$ Oxygen left unreacted

$=100-52.5=47.5ml$

$x+z=47.5ml$

$x=(47.5-z)$ -------- Eqn. $2$

Substitute $2$ in $1$

$47.5-z=y-2z$

$y-z=47.5ml$ ------ Eqn. $3$

Again, since

$x+y+z=100$

$(47.5-z)+y+z=100$

$\Rightarrow y=52.5ml$

$\Rightarrow z=52.5-47.5=5ml$

$\Rightarrow x=100-(52.5+5)=42.5ml$

$V_{CO}:V_{H_2}:V_{C{O}_2}$

$\frac{42.5}{2.5}=\frac{52.5}{2.5}=\frac{5}{2.5}$

$17:21:2$

$1721$