I'm not exactly sure what the problem does but I think the idea is to show that the only way we can obtain 55.4% CuO is this way.
The mass of the reactants is 9.56 g + 4.08 g = 14.36 g.
The Law of Conservation of Mass says that the products must also have a mass of 14.36 g. So how much CuO would form if we had 55.4% CuO? It would be (x/0.554) = 14.36 or
0.554 x 14.36 = 7.95 g.CuO
The other product is SO2 and the mass of that will be 14.36 x 0.446 = 6.41 g
I did not understand the lesson today in my Chemistry I honors class.
We went over the Law of Conservation of Mass for roughly 15 minutes and he gave us this worksheet. Could someone please display how to do this, I'll pull a problem from the worksheet.
When 9.56g of Copper Sulfide (CuS) is heated in the presence of 4.80g of Oxygen (O), the products contain 55.4% Copper Oxide. What is the total mass of products? What mass of copper oxide is formed?
My equations so far looks similar to this:
CuS + O -----> CuO 55.4%
9.56g .36g xg
2 answers
Very helpful, thank you. I will try this method on other similar problems, and yes, I was not sure what the whole point of the problem is. Thanks once again!