moles O2 lost = 0.302g/molar mass O2 = about 0.009 (that's an estimate).
Convert mols O2 to mols Mg(ClO3)2 with the coefficients.
0.009 x (2 mol Mg(ClO3)2/3 mols O2) = 0.009 x 2/3 = about 0.006
g Mg(ClO3)2 = mols x molar mass
Then % Mg(ClO3)2 = [g Mg(ClO3)2/mass sample)*100 = ?
Write the balanced equation given that Mg(ClO3)2 decomposes into chloride and oxygen:
2Mg(ClO3)2 --> 2MgCl2 + 3O2
So if 0.302 grams of O2 is lost from 1.890 grams of a mixture of Mg(ClO3)2 and an inert material, what is the percentage of Mg(ClO3)2 in the mixture?
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