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

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

A

1 mole of `HCl_((g))`

B

2 mole of `HCl_((g))`

C

0.5 mole of `HCl_((g))`

D

1.5 mole of `HCl_((g))`

Text Solution

Verified by Experts

The correct Answer is:
A, C
`H_(2)+Cl_(2) rarr 2HCl`
`22.4 "lt. " 11.2 "lt."`
1 moleof `HCl=(1)/(2)` mole
Limiting reagent is `Cl_(2)`. So, 1 mole HCl is formed.
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