Answer:
2.2 grams
Explanation:
2KCl + 3O2 --> 2KClO3
KCl = 39+ 35.5
= 74.5 g/mol
O = 16 g/mol so O2 is 32 g/mol
We need 3 mols of O2 for every 2 mols of KCl
50 g KCl = 50/74.5 = .671 mols of KCl
we need .671 * 3/2 = 1.007 mols of O2 = 32 g
Now we have more O2 than we need, just use the .671 mols of KCl, limiting reagent
now
We get 1 mol KClO3 for every mol of KCl
then we get .671 mols of KClO3
74.5 + 3*16 = 122.5 grams/mol of KClO3
.671 * 122.5 = 2.2 grams
