Given the reaction :

A_{2_{(g)}}\; + \; xB_{(g)}\; \rightarrow\; A_{2}B_{x}

If the volumes occupied by A and B at STP are 22.4L and 44.8L, then the value of x is ?

Option 1)  2

Option 2)  4

Option 3)  3

Option 4)  1

Answers (1)

As we learnt.

Gay Lussac's law of Gaseous Volumes -

When gases combine or are poduced in a chemical reaction, they do so in a simple ratio by volume provided all gases are at the same temperature and pressure.

- wherein

H_{2}}_{(g)} + {O_{2}}_{(g)} \rightarrow {H_{2}O}_{(g)}

100 ml        50ml            100ml

 

By Gay Lussac's Law, volume occupied by gases in a reaction are in simple ratio, that is fixed.

Moles of A_{2} = \frac{22.4}{22.4} = 1 mole

Moles of B = \frac{44.8}{22.4} = 2 mole

\therefore x = 2  

\therefore Solution is A

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