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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 L of `Cl_(2)(g)`, each at STP, the moles of HCl(g) formed is equal to

A

0.5 mol of HCI (g)

B

1.5 mol of HCI (g)

C

1 mol of HCI (g)

D

2 mol of HCI (g)

Text Solution

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The correct Answer is:
A
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