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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: Aaa on February 05, 2015, 01:29:26 AM

Title: Gas stoichiometry
Post by: Aaa on February 05, 2015, 01:29:26 AM
CaO, is produced by heating calcium carbonate, CaCO3. Calculate the volume of CO2 expected to be produced, at STP, from the complete decomposition of 152.0 grams of calcium carbonate.
CaCO3(s) :rarrow: CaO(s)+ CO2(g)
152.0g CaCO3 × 1mol CaCO3 x 1ml CO2 = 15.07 mol CO2
                  100.09g CaCO3  1mol CaCO3
V= nRT/P

V= (15.07mol)(0.0821atm·L/mol·k)(273K)   
                   1atm
ans: 338L CO2 (g)

I'm not sure if i'm doing this right because i believe that i'm supposed to find the theoretical yield of CO2 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!!~
Title: Re: Gas stoichiometry
Post by: sjb on February 05, 2015, 02:03:25 AM
Quote from: Aimmer on February 05, 2015, 01:29:26 AM
CaO, is produced by heating calcium carbonate, CaCO3. Calculate the volume of CO2 expected to be produced, at STP, from the complete decomposition of 152.0 grams of calcium carbonate.
CaCO3(s) :rarrow: CaO(s)+ CO2(g)
152.0g CaCO3 × 1mol CaCO3 x 1ml CO2 = 15.07 mol CO2
                  100.09g CaCO3  1mol CaCO3
V= nRT/P

V= (15.07mol)(0.0821atm·L/mol·k)(273K)   
                   1atm
ans: 338L CO2 (g)

I'm not sure if i'm doing this right because i believe that i'm supposed to find the theoretical yield of CO2 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!!~

Your calcium carbonate is your limiting reagent here, so that's all fine. Check the number of moles of it that you have though.
Title: Re: Gas stoichiometry
Post by: AWK on February 06, 2015, 06:10:27 AM
152 g CaCO3 = 1.52 mole