Question

Calculate the volume of hydrogen required for complete hydrogenation of $0.25d{m}^3$ of ethyne at $STP$.

Answer 1

However, an interesting thing takes places when you are dealing with gases that are under the same conditions for pressure and temperature. in such cases the mole ratio becomes the volume ratio.

the idea is that since you are dealing with two ideal gases. you can use the ideal gas law equation to write

$P.V_1=n_1.RT$ (for ethyne)

$P.V_2=n_2.RT$ (for hydrogen)

The pressure and the temperature are the same for both gases. since the reaction presumably takes place ta STP.

if you divide these two equal ions, you will get.

$\frac{P.V_1}{P.V_2}=\frac{n_1.RT}{n_{2}RT}$

this is equivalent to $\frac{n_1}{n_2}=\frac{V_1}{V_2}$

the mole ratio is equivalent to the volume ratio.

All you need now is the balanced chemical equation for the reaction which looks like this

$C_2{H}_2\left(g\right)+2H_2\left(g\right)\to C_2{H}_6\left(g\right)$

Notice, that you have 1:2 mole ratio between ethyne and hydrogen gas. this means that the reaction needs twice as many moles of hydrogen gas than of ethyne $0.25d{m}^{3}ethyne=\frac{2d{m}^{3}hydrogen}{1d{m}^{3}ethyne}$

$0.50d{m}^3$ hydrogen.