Question

Consider the following reaction between calcium oxide and carbon dioxide: Cao (0) + CO2(g) Cacos@) A chemist allows 14.4 g of Cao and 13.8 g of CO2 to react. When the reaction is finished, the chemist collects 19.9 g of Caco, Part A Determine the limiting reactant for the reaction Express your answer as a chernkal formula. .- AEGO ? Submit Request Answer

Answer 1

CaO(s) + CO2(g) -------> CaCO3(s)
no of moles of CaO = W/G.M.Wt
                    = 14.4/56   = 0.257moles
no of moles of CO2 = W/G.M.Wt
                    = 13.8/44 = 0.3136moles
CaO(s) + CO2(g) -------> CaCO3(s)
1 mole of CO2 react with 1 mole of CaO
0.3136 moles of CO2 react with 0.3136 moles of Cao is required
CaO is limiting reactant >>>>answer
1 mole of CaO react with excess of CO2 to gives 1 moles of CaCO3
0.257 moles of CaO react with excess of CO2 to gives 0.257moles of CaCO3
mass ofCaCO3 = no of moles * gram molar mass
              = 0.257*100
              = 25.7g
Theoretical yield = 25.7g
actual yield of CaCO3 = 19.9g
percent yield = actual yield*100/Theoretical yield
              = 19.9*100/25.7 = 77.43%