Answer to Question #128272 in General Chemistry for Pokemon

Question

Consider the reaction between calcium oxide and carbon dioxide.
CaO + CO2>>> CaCO3
A chemist allows 14.4g of CaO and 13.8g of CO2 to react. When the reaction is finished the chemist collects 19.4 of CaCO3. Determine the limiting reactant, theoritical yield and precent yield for the reaction.
(Molar masses, g/mol:CaO=56.08,CO2=28.01,CaCO3=100.09)

Answer 1

from the number of moles of reactants and products, we can determine the % yield and thus establish the limiting reactant

number of moles is calculated from the ratio of mass and molar mass

number of moles of CaO = 14.4/56.1= 0.26 mol

nummber of moles of CO₂= 13.8/44= 0.31 mol

to establish the limiting reactant, compare the number of moles of CaO, CO₂ with that of CaCO₃ produced :

the reacting mole ratios of reactants and products = 1:1:1

hence, the number of moles of CaCO₃=0.26 mol

The number of moles of CaCO₃ produced by CaO is less than the theoretical yield, hence CaO is the limiting reactant: if amount of CaO is increased then more of CaCO₃ will be produced from the available CO₂

mass of CaCO₃ produced = 0.26*100=26 g where 100 = molar mass of CaCO₃

% yield = (actual yield/theoretical yield)*100

= (19.4/26)*100

=74.6%