Did you look at my diagram?
"CaCO3 --> CaO + CO2
and related to that question, what mass in grams of calcium oxide is produced from 50g of calcium carbonate?"
the mole ratio is 1:1:1 (you understand that yes?) Ok, so going from the diagram, we have
Grams A (thats the CaCO3) = 50 g.. and we cant go straight to grams B. So now we have to work out the how many moles of CaCO3 in 50 grams. Thats where the formula comes in. n=m/M
n = 50/100.09 (sum mole weights of Ca = 40.08, C= 12.01 and 3xO =16)
n = 0.4996....
now we have moles A. Since the mole ratio of CaCO3:CaO is 1:1, then moles B is also 0.4996...
to get back to grams B, we have to use that formula again.
n = m/M
0.4996... = m/56.08 (sum mole weights of Ca = 40.08, O = 16)
m = 0.4996 x 56.08
m = 28.02 g
So, approx 28.02 grams produced. Not exact cos i rounded the .4996 off early. And i presume you understand how to get the mole weight of a substance -- it is just the weights for the elements on the periodic table.
Tell me if that does or doesnt all make sense to you. If its the wrong answer its most likely a calculation error... that is the method you should use. If its hard to follow i will write my working and upload it for you later. Ill be away this weekend so i hope you dont need any more urgent help =p