A5.00g 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?

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!