The equation is right.
Mg(ClO3)2 --> MgCl2 + 3O2
Convert g O2 to mols. mols = grams/molar mass = about(estimated)0.009 but you should do it more accurately.
Convert mols O2 to mols Mg(ClO3)2 using the coefficients in the balanced equation. That is
0.009 x (1 mol Mg(ClO3)2/3 mols O2) = 0.009 x (1/3) = about 0.003 (again an estimate).
Convert mols Mg(ClO3)2 to g. g = mols x molar mass
Then %Mg(ClO3)2 = (g Mg(ClO3)2/mass sample)*100 = ?
Given that Mg(ClO3)2 decomposes into magnesium chloride and oxygen:
A) Write a balanced equation for this reaction.
I got Mg(ClO3)2 yields MgCl2 + 3 O2
Is that right?
B) 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?
Please help! I'm lost.
1 answer