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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

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
`{:(,,H_(2),+,Cl_(2),to,2HCl),("Intial",:,22.4L,,11.2L,,0),("Final",:,11.2L,,0,,22.5L=1"mole"):}`
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