1.

5.0 g of marble was added to 7.5 g dilute hydrochloric acid. After the reaction was over, it was found that 0.5 g of marble was left unused. Calculate the percentage strength of hydrochloric acid. What volume of CO_(2) measured at STP will be evolved in the above reaction?

Answer»


Solution :`underset(100g)(CaCO_(3))+underset(2xx36.5g=73g)(2HCl)rarrCaCl_(2)+H_(2)O+underset(22400cm^(3)" at STP")(CO_(2))`
`"MARBLE reacted "=5-0.5 G=4.5g`
`"HCl reacted with 4.5 g marble"=(73)/(100)xx4.5g=3.285g`
`%" strength"=(3.285)/(7.5)xx100=43.8%`
`CO_(2)" evolved at S.T.P."=(22400)/(100)xx4.5cm^(3)=1008cm^(3)`


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