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Calculate the volume of H(2) gas (at STP...

Calculate the volume of `H_(2)` gas (at STP) liberated by the reaction of exces Zn with 500 mL 0.5(N) `H_(2)SO_(4)`.

Text Solution

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Reaction involved: `Zn+underset(98g)(H_(2)SO_(4)) to ZnSO_(4)+underset(22400 mL)(H_(2))`
`H_(2)SO_(4)` present in 1000 mL of 0.5(N) `H_(2)SO_(4)=0.5` g-eqv.
`therefore`Amount of `H_(2)SO_(4)` present in 500 mL of 0.5(N) `H_(2)SO_(4)`
solution`=(0.5xx500)/(1000)` g -equiv.`=(0.5xx500xx49)/(1000)g=12.25g`
98g `H_(2)SO_(4)` reacts with Zn to liberated 22400 mL `H_(2)` (STP).
`therefore 12.25g" "H_(2)SO_(4)` reacts with zinc to liberate
`=(22400xx12.25)/(98)=2800mL" "H_(2)` (at STP)
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