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When 22.4 L of H(2)(g) is mixed with 11....

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

A

2 moles of HCI (g)

B

0.5 moles of HCI (g)

C

1.5 moles of HCI (g)

D

1 moles of HCI (g)

Text Solution

Verified by Experts

The correct Answer is:
A, C
`H_2(g)+CI_(2)(g) to 2HCI`
1 mole of an ideal gas occupies at 22.4L.
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