$$10$$ grams of limestone on calcination liberates $$1.68$$ lit of $$CO2$$ at STP. Percentage purity of limestone is

Question

$$10$$ grams of limestone on calcination liberates $$1.68$$ lit  of $$CO2$$ at STP. Percentage purity of limestone is

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$$CaCO3S$$ →   CaOs  $$+$$  $$CO2g100$$ g                                                     $$22.4$$ $$lit10$$ g                                                        $$2.24$$ lit    $$10$$ grams of $$100$$% pure limestone liberates $$2.24$$ lit $$CO2$$ of  at $$STP10$$ grams of “X%” pure limestone liberates $$1.68$$ lit of $$CO2$$  at STPPercentage purity  (X)  $$=$$ $$1.682.24$$  ×  $$100$$                                                $$=$$ $$75$$ %

Stoichiometry and Stoichiometric Calculations

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10 grams of limestone on calcination liberates 1.68 lit of ${CO}_{2}$ at STP. Percentage purity of limestone is

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10 grams of limestone on calcination liberates 1.68 lit of CO2 at STP. Percentage purity of limestone is

CaCO3S → CaOs + CO2g100 g 22.4 lit10 g 2.24 lit 10 grams of 100% pure limestone liberates 2.24 lit CO2 of at STP10 grams of “X%” pure limestone liberates 1.68 lit of CO2 at STPPercentage purity (X) = 1.682.24 × 100 = 75 %

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10 grams of limestone on calcination liberates 1.68 lit of ${CO}_{2}$ at STP. Percentage purity of limestone is

${CaCO}_{3} (S) \overset{}{\to} CaO (s) + {CO}_{2} (g)$
100 g 22.4 lit
10 g 2.24 lit
10 grams of 100% pure limestone liberates 2.24 lit ${CO}_{2}$ of at STP
10 grams of “X%” pure limestone liberates 1.68 lit of ${CO}_{2}$ at STP
Percentage purity (X) = $\frac{1 .68}{2 .24} \times 100$
= 75 %