Asked by ATS
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
Answered by
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
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