Lessons

Eamcet Chemistry
1st Year : Unit 1.4 (Basics & Concepts) - Stoichiometric Calculations

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Q.

How many grams of 80% pure marble stone on calcinations can give 14 grams of quick lime?

View Answer

Correct choice: C

Explanation:

CaCO3 → CaO + CO2 1 mole of CaO = 1 mole of CaCO3 56 grams of CaO given by 100 grams of CaCO3 14 grams of CaO given by ( 14/56)100=25 grams of CaCO3 As purity is 80%
i.e 80 grams of calcium carbonate is present in 100 grams of marble stone
25 grams of calcium carbonate is present in (25X100 /80)=31.25grams of marble stone
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