A sample of a mixture of CaCl2 and Na2CO3 weighing 4.22g was treated to precipitate all the Ca as CaCO3. This CaCO3 is heated and quantitatively converted into 0.959g of CaO. Calculate the percentage of CaCl2 in the mixture. (Atomic mass of Ca=40, O=16, C=12 and Cl=35.5 g/mole)

CaCO3 → CaO + CO2

The molar mass of CaO is 40 + 16 = 56 g/mol.

The number of moles of CaO = mass / molar mass = 0.959 g / 56 g/mol = 0.017125 mol

Since 1 mole of CaCO3 produces 1 mole of CaO, the number of moles of CaCO3 is also 0.017125 mol.

The molar mass of CaCO3 is 40 + 12 + (3 * 16) = 100 g/mol.

The mass of CaCO3 = number of moles * molar mass = 0.017125 mol * 100 g/mol = 1.7125 g

Since all the calcium in the mixture is precipitated as CaCO3, the mass of Ca in CaCO3 is:

Mass of Ca = (40/100) * 1.7125 g = 0.685 g

The molar mass of CaCl2 is 40 + (2 * 35.5) = 111 g/mol.

The number of moles of Ca in CaCl2 is the same as the number of moles of Ca in CaCO3, which is 0.017125 mol (since 1 mole CaCl2 produces 1 mole Ca).

Mass of CaCl2 = number of moles * molar mass = 0.017125 mol * 111 g/mol = 1.90 g

Percentage of CaCl2 in the mixture = (mass of CaCl2 / total mass of mixture) * 100

Percentage of CaCl2 = (1.90 g / 4.22 g) * 100 = 45.02%

Therefore, the percentage of CaCl2 in the mixture is approximately 45.00%