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Calculate the volume of hydrogen liberated at STP when 500 cm^(3) of sulphuric acid reacts with excess of zinc. (H = 1, O = 16, S = 32) |
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Answer» Solution :The chemical equation representing the reaction is : `Zn+underset(98.0g)(H_(2)SO_(4))rarrZnSO_(4)+underset(22400cm^(3)"at STP")(H_(2))` STEP 1. To calculate the AMOUNT of `H_(2)SO_(4)` in `500 cm^(3)` of 0.5 N `H_(2)SO_(4)` solution. Strength/litre of `H_(2)SO_(4)"solution" = "Normality"xx"Eq. mass"=0.5xx49=24.5"g/litre"` Now, `1000 cm^(3)` of the acid solution contain 24.5 g of pure `H_(2)SO_(4)` `therefore 500cm^(3)" of the acid solution will contain pure "H_(2)SO_(4)=24.5xx(500)/(1000)=12.25g` Step 2. to calculate the volume of `H_(2)` liberated at STP `" 98 g of "H_(2)SO_(4)" react to liberate "22400cm^(3)" of "H_(2)" at STP"` `therefore"12.25 g of "H_(2)SO_(4)" would liberate H"_(2)" at STP"=(22400)/(98)xx12.25cm^(3)=2800cm^(3)` Thus, the volume of HYDROGEN liberated at STP = `2800cm^(3)` |
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