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%
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