Can someone please help with this chem question

Question

1. A mixture of calcium chloride and calcium chlorate was treated according to the procedure used in this exercise. An initial mass of 1.093g of the salt mixture, brought to a constant mass after heating, was found to have lost 0.219g. Determine the mass percent of calcium chlorate present in the original sample.

Answer 1

You have calcium chloride, CaCl₂, and calcium chlorate, Ca(ClO₃)₂

You heat the mixture and essentially nothing happens to the CaCl2, but the Ca(ClO3)2 decomposes

Ca(ClO₃)₂ ==> CaCl₂ + 3O₂ .. balanced equation for decomposition of Ca(ClO3)2

The O₂ gas that is formed will be lost and that is the loss in mass that you see (0.219 g)

Mass lost = 0.219 g. This is the mass of O₂ in the above equation. This can come only from the Ca(ClO₃)₂, and not from the CaCl₂.

Now we can fine how much Ca(ClO₃)₂was present.

0.219 g O₂ x 1 mol O₂ / 32 g = 0.00684 moles O₂ lost

0.00684 mols O₂ x 1 mol Ca(ClO₃)₂ / 3 mols O₂ = 0.00228 mols Ca(ClO₃)₂ originally present

molar mass of Ca(ClO₃)₂ = 207 g / mol

mass of Ca(ClO₃)₂ originally present = 0.00228 mols x 207 g / mol = 0.472 g Ca(ClO₃)₂ originally present

% Ca(ClO₃)₂ = 0.472 g / 1.093 g (x100%) = 43.2%