Answer :
Final answer:
The percentage mass of CaCO₃ in the limestone is 90.34%, calculated by finding the number of moles of CO₂ produced, converting these into moles and then grams of CaCO₃, and comparing this mass to the total mass of the sample.
Explanation:
A given sample of limestone has produced 0.637g of CO₂ when heated. In order to find out what percentage this represents in the sample of limestone, you need to know the molar mass of both CO₂ and CaCO₃. These are approximately 44.01g/mol and 100.09g/mol respectively.
First, calculate the number of moles of CO₂: (0.637g/44.01g/mol) = 0.0145 mol. There is a 1:1 stoichiometric ratio between CO₂ and CaCO₃, so the number of moles of CaCO₃ are the same: 0.0145 mol.
Next, convert the moles of CaCO₃ into grams: (0.0145 mol * 100.09g/mol) = 1.451g. Now, find the percentage mass: (mass of CaCO₃/mass of sample) * 100% = (1.451g / 1.605g) * 100% = 90.34%
Therefore, the percentage mass of CaCO₃ in the limestone is 90.34%.
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