CaO, is produced by heating calcium carbonate, CaCO
3. Calculate the volume of CO
2 expected to be produced, at STP, from the complete decomposition of 152.0 grams of calcium carbonate.
CaCO
3(s)
CaO(s)+ CO
2(g)
152.0g CaCO
3 ×
1mol CaCO3 x
1ml CO2 = 15.07 mol CO
2 100.09g CaCO
3 1mol CaCO
3V= nRT/P
V=
(15.07mol)(0.0821atm·L/mol·k)(273K) 1atm
ans: 338L CO
2 (g)
I'm not sure if i'm doing this right because i believe that i'm supposed to find the theoretical yield of CO
2 because the problem says 'expected to be produced'. But to find TY i'm supposed to have a limiting reagent and an excess?? So.. what i'm trying to ask is how do you find the limiting reagent and excess of a decomposition formula? I dunno i'm really confused!!~