Q.

The weight of CaCo3 needed for the preparation of 2.24 lit of Acetylene gas under S.T.P conditions is

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Detailed Solution

CaC21 mole+2H2O→CaOH2+C2H21 mole 1 mole CaC2 on hydrolysis gives 1 mole acetylene,1 mole CaCO3 produces 1 mole of CaC2.CaCO31 mole100 gr→CaC21 mole →C2H21 mole22.4 L      ??                        2.24L      2.24×10022.4=10 gr
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