1.

10 g of a piece of marble was put into excess of dilute HCl acid. When the reaction was complete, 1120 cm^(3)of CO_(2)was obtained at STP. The percentage of CaCO_(3)in the marble is

Answer»

`25%`
`50%`
`75%`
`100%`

Solution :`underset(100 G)(CaCO_(3)) + 2HCL to CaCl_(2) + H_(2)O + CO_(2)`
1120 cc of `CO_(2)` are obtained from pure `CaCO_(3)``=100/22400 xx 1120 = 50 g`
% of `CaCO_(3) = 5/10 xx 100 = 50%`


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