1.

Calculate the weight of 80% pure limestone required to produce 11 g of CO_(2) gas.

Answer»

Solution :`{:(CaCO_(3) to CaO+CO_(2)),(" 1 MOLE""1 mole"):}`
44 g of `CO_(2)` is produced from 100 g of `CaCO_(3)`
`:. 11" g of "CO_(2)` is produced from` 100/44 xx 11 = 25" g of "CaCO_(3)`
As thelimestone is 80% pure, THEWEIGHT of IMPURE LIMESTONE required = `100/80 xx 25" g " = 31.2` g.


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