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When 22.4 litres of H(2(g)) is mixed wit...

When 22.4 litres of `H_(2(g))` is mixed with 11.2 litres of `Cl_(2(g))` each at S.T.P, the moles of `HCl_((g))` formed is equal to

A

1 mol of `HCl_((g))`

B

`2 mol " of " HCl_((g))`

C

`0.5 ` mol of ` HCl_((g))`

D

1.5 mol of ` HCl_((g))`

Text Solution

Verified by Experts

The correct Answer is:
A
1 mole = 22.4 litres at S.T.P.

Here, `Cl_2` is limiting reagent. So, 1 mole of `HCl_((g))` is formed.
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