Question

A 5.00g sample of impure CaCO3 is found to react completely with 100.0 mL of 0.100 M of H2SO4. The equation for the reaction is: CaCO3 (s) + H2SO4 (aq) —> CaSO4 (s) + CO2 (g) + H2O (l). What is the mass percent CaCO3 in the impure sample?

Answer 1

Hey there!:

Given the reaction:

CaCO3(s) + H2SO4(aq) -------> CaSO4(s) + CO2(g) + H2O(l)

volume in liters( H2SO4 ) = 100 mL / 1000 => 0.1 L

Moles of H2SO4:

Molarity * Volume ( L )

0.100 * 0.1 => 0.01 moles of H2O4

1 mole H2SO4 reacted with -------------- 1 mole CaCO3

0.01 moles H2SO4 reacted with ------- 0.01 moles CaCO3

Molar mass CaCO3 = 100.0869 g/mol

Mass of CaCO3 :

Moles of CaCO3 * Molar mass CaCO3

0.01 * 100.0869 => 1.000869 g of CaCO3

Therefore:

% mass = ( Mass of CaCO3 / mass of sample ) * 100

% mass = ( 1.000869 / 5.00 ) * 100

% mass = 0.2001738 * 100

% mass = 20.01%

Hope that helps!