When a 1.25-gram sample of limestone was dissolved in acid, 0.44 gram of CO2 was generated. If the rock contained no carbonate other than CaCO3, what was the percent of CaCO3 by mass in the limestone?

80%

Explanation:

First off, we write out the chemical equation for the reaction, assuming the acid used is HCl;

CaCO3 + 2 HCl → CaCl2 + H2O + CO2

1 mol of CaCO3 generates 1 mol of CO2

Molar mass of CaCO3 = 40 + 12 + (3 * 16) = 52 + 48 = 100 g/mol

Molar mass of CO2 = 12 + (2*16) = 12 + 32 = 44 g/mol

So mass of CaCO3 = 1 mol * 100 g/mol = 100 g

Mass of CO2 formed = 1 mol * 44 g/mol = 44 g

This means 100g of CaCO3 generated 44g of CO2

How much would then generate 0.44g?

100 = 44

x = 0.44

Upon cross Multiplication we have;

x = (0.44 * 100) / 44

x = 1g

Percent by mass of CaCO3 = (mass of CaCO3 / Mass of Limestone) * 100

Percent by mass of CaCO3 = (1 / 1.25) * 100 = 0.8 * 100 = 80%