July 03, 2020, 03:36:32 PM
Forum Rules: Read This Before Posting


Topic: Gas stoichiometry  (Read 1295 times)

0 Members and 1 Guest are viewing this topic.

Offline Aaa

  • Very New Member
  • *
  • Posts: 1
  • Mole Snacks: +0/-0
Gas stoichiometry
« 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!!~

Offline sjb

  • Global Moderator
  • Sr. Member
  • ***
  • Posts: 3464
  • Mole Snacks: +217/-42
  • Gender: Male
Re: Gas stoichiometry
« Reply #1 on: February 05, 2015, 02:03:25 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.

Offline AWK

  • Retired Staff
  • Sr. Member
  • *
  • Posts: 7374
  • Mole Snacks: +516/-86
  • Gender: Male
Re: Gas stoichiometry
« Reply #2 on: February 06, 2015, 06:10:27 AM »
152 g CaCO3 = 1.52 mole
AWK

Sponsored Links