A mixture of salt way in 13.78g contains CACO3 with gaseous impurity it was heated to a constant mass and the resolve was reacted with H2SO4 to form 6.78g of CASO4. Calculate the percentage purity of the calcium trizoncarbonate four I'm the mixture (ca=40, c=12, H=1, O=16, sulphur=32)
Equation of the reaction cao
1. Calcium oxide + H2SO4→CASO4 + H2O
2. CaCo3→CaO2 + O2
To calculate the percentage purity of calcium trizoncarbonate (CaCO3) in the mixture, we need to determine the amount of CaCO3 present and compare it to the total mass of the mixture.
1. Calculate the molar mass of CaSO4:
Ca = 40 g/mol
S = 32 g/mol
O = 16 g/mol (x4 in the formula)
Molar mass of CaSO4 = (40 + 32 + (16 * 4)) g/mol = 136 g/mol
2. Calculate the moles of CaSO4 formed:
Moles of CaSO4 = Mass / Molar mass = 6.78 g / 136 g/mol = 0.05 mol
3. Determine the stoichiometry of the reaction between CaCO3 and CaSO4:
From the balanced equation: CaCO3 → CaO + CO2
The stoichiometry is 1:1, meaning 1 mole of CaCO3 reacts to give 1 mole of CaSO4.
4. Calculate the moles of CaCO3 in the mixture:
Moles of CaCO3 = Moles of CaSO4 = 0.05 mol
5. Calculate the mass of CaCO3 in the mixture:
Mass of CaCO3 = Moles of CaCO3 × Molar mass of CaCO3
Molar mass of CaCO3 = (40 + 12 + (16 * 3)) g/mol = 100 g/mol
Mass of CaCO3 = 0.05 mol × 100 g/mol = 5 g
6. Calculate the percentage purity of CaCO3:
Percentage purity = (Mass of pure CaCO3 / Total mass of the mixture) × 100
Percentage purity = (5 g / 13.78 g) × 100 = 36.28%
Therefore, the percentage purity of the calcium trizoncarbonate (CaCO3) in the mixture is approximately 36.28%.