Question
An antacid tablet containing calcium carbonate as an active ingredient required 9.19 mL of 0.0956 M H2SO4 for complete neutralization. The mass of the tablet was 0.193 g. What percent of calcium carbonate was in the tablet? Thanks!
Answers
Steve
9.19mL of .0956M H2SO4 contains
.00919*.0956 = .0008786 moles H2SO4
CaCO3 + H2SO4 = CaSO4 + H2CO3
since each mole of H2SO4 racats with one mole CaCO3, there were .0008786 moles CaCO3 in the tablet.
Since each mole CaCO3 has mass 100g, the pill was .08786/.193 = .455 = 45.5% CaCO3
.00919*.0956 = .0008786 moles H2SO4
CaCO3 + H2SO4 = CaSO4 + H2CO3
since each mole of H2SO4 racats with one mole CaCO3, there were .0008786 moles CaCO3 in the tablet.
Since each mole CaCO3 has mass 100g, the pill was .08786/.193 = .455 = 45.5% CaCO3