Okay, so here's the question:
"In the lab, the salt potassium perchlorate, KClO4, can be decomposed thermally to general small amounts of oxygen gas. 2KClO4(s) -> 2KCl + 3O2(g) How many grams of potassium perchlorate must be decomposed to produce 12.89g of O2?"
Can anyone tell me how you solve this problem? I mean, sure we went over thermal reactions in lecture, but we never had any problems like this on any quiz of exam. To be honest, I don't think there's enough information here to actually solve this problem...at least I don't think there's enough information. Any chem experts willing to share their knowledge with me?
"In the lab, the salt potassium perchlorate, KClO4, can be decomposed thermally to general small amounts of oxygen gas. 2KClO4(s) -> 2KCl + 3O2(g) How many grams of potassium perchlorate must be decomposed to produce 12.89g of O2?"
Can anyone tell me how you solve this problem? I mean, sure we went over thermal reactions in lecture, but we never had any problems like this on any quiz of exam. To be honest, I don't think there's enough information here to actually solve this problem...at least I don't think there's enough information. Any chem experts willing to share their knowledge with me?
-
The equation you have given, using potassium perchlorate, isn't correctly balanced.
The left side has more oxygen atoms than the right side does. When I did this in
the lab, we used potassium chlorate (KClO3) and not the perchlorate. If you use
potassium chlorate, the equation balances: 2KClO3 → 2KCl + 3O2(g)
The balanced equation shows that 2 molecules of potassium chlorate produce 3
molecules of oxygen (O2) gas. So we need two-thirds as many moles of chlorate
as we have moles of oxygen produced. The molecular weight of oxygen is 32, so
we produce 12.89 grams divided by 32 gm. per mole, or 0.4028 mole of O2 gas.
So we will need 2/3 that number, or 0.2685 mole, of potassium chlorate.
The molecular weight of potassium chlorate is 122.55 (grams per mole) and we
need 0.2685 mole, so we need 122.55 times 0.2685, or 32.9 grams of KClO3.
So you do have enough information to solve this problem. Check with your
teacher to be sure it's potassium chlorate, so the equation balances correctly.
The left side has more oxygen atoms than the right side does. When I did this in
the lab, we used potassium chlorate (KClO3) and not the perchlorate. If you use
potassium chlorate, the equation balances: 2KClO3 → 2KCl + 3O2(g)
The balanced equation shows that 2 molecules of potassium chlorate produce 3
molecules of oxygen (O2) gas. So we need two-thirds as many moles of chlorate
as we have moles of oxygen produced. The molecular weight of oxygen is 32, so
we produce 12.89 grams divided by 32 gm. per mole, or 0.4028 mole of O2 gas.
So we will need 2/3 that number, or 0.2685 mole, of potassium chlorate.
The molecular weight of potassium chlorate is 122.55 (grams per mole) and we
need 0.2685 mole, so we need 122.55 times 0.2685, or 32.9 grams of KClO3.
So you do have enough information to solve this problem. Check with your
teacher to be sure it's potassium chlorate, so the equation balances correctly.