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%