Question

Gas analysis of sample shows that it has 20% H2, 40% CH4, 10% CO, 20% C2H6 and 10% noncombustible
inert gases (Volume%). 2m³ volume of above sample is combusted (burnt) completely with oxygen. If X is
the volume of air required (in m³). Calculate the value of (10X/33).(Air consist 20% oxygen by volume)​

Answer 1

Answer:

The answer is 5

Explanation:

The explanation is in attachment.

HOPE IT HELPS..

Answer 2

$2H_{2}+O_{2}- > 2H_{2}O$

$CH_{4}+2O_{2}- > CO_{2}+2H_{2}O$

$2CO+O_{2}- > 2CO_{2}$

$C_{2} H_{6}+\frac{7}{2}O_{2}- > 2CO_{2}$

(2×0.2)$m^{3}$     $\frac{7}{2}$×(2×0.2)$m^{3}$+$3H_{2}O$

$V_{O_{2} }$$=[\frac{2*0.2}{2}+2*(2*0.4)+\frac{2*0.1}{2}+\frac{7}{2}(2*0.2)]m^{3}$

$=[0.2+1.6+1.4]m^{3}$

$=3.3m^{3}$

Let the volume of air needed be x.

Therefore, the volume of oxygen in it is,

$=\frac{20}{100}*x=0.2x$

$0.2x=3.3m^{3}$

$x=\frac{33}{2}$ $m^{3}$

That is,

$\frac{10x}{33}=\frac{10}{33}*\frac{33}{2}=5$

So the value of $\frac{10x}{33}$ is 5.