Chapter 3, Problem 83A

Interpretation Introduction

Interpretation:

The amount of hydrogen reacted and thetotal amount of hydrogen that was used initially needs to be determined.

Concept introduction:

Gay Lussac’s Law explains that when a reaction goes all the reactants react in a definite volumes and produces the products in definite volume also.

Answer to Problem 83A

The initial mass of hydrogen present before the reaction was $(11+10.5)\mathrm{gm}=21.5gm$.

Mass of hydrogen used in the reaction is $10.5gm$.

Explanation of Solution

As per the balanced chemical equation:

$2P+3H_2\to 2P{H}_3$

Two moles of phosphorus reacts with three moles of hydrogen to form two moles of phosphine.

If $108.3\mathrm{gm}$ of phosphorus is reacted with excess of hydrogen the amount of phosphine produced is $129.9\mathrm{gm}$. At the end of reaction, $11\mathrm{gm}$ of hydrogen remain unreacted.

$2P+3H_2\to 2P{H}_3$

Here $62gm$ of phosphorus reacts with $6gm$ of hydrogen to form $68gm$ of phosphine.

But it is $108.3\mathrm{gm}$ phosphorus that produces $129.9\mathrm{gm}$ of phosphine.

1. Making equation between phosphorus and hydrogen get:

$\begin{array}{l}62gmreacts\to 6gm\\ 108.3gmwillreact\to ?(x)\\ x=\frac{108.3\times 6}{62}=10.5gm\end{array}$

So $108.3\mathrm{gm}$ of phosphorus reacts $10.5gm$ of hydrogen.

2. Making equation between Phosphine and hydrogen :

$\begin{array}{l}6gm{H}_{2}gives\to 68{\mathrm{gmPH}}_3\\ (y)gm{H}_{2}willgive\to 129.9gmP{H}_3\\ y=\frac{129.9\times 6}{68}=11.5\mathrm{gm}\end{array}$

For getting $129.9\mathrm{gm}$ of phosphine we need to use $11.5\mathrm{gm}$ of hydrogen which isn’t used in the last equation.

Conclusion

When gases react they do so in definite volume as per Gay Lussac’s Law. So here also the number of particles of hydrogen should be double to the number of particles of oxygen.