Question

Consider the reaction between calcium oxide and carbon dioxide:  

            CaO (s) + CO2(g)  →  CaCO3(s)

A chemist allows 15.27 g of CaO and 18.06 g of CO2to react. When the reaction is finished, the chemist collects 22.15 g of CaCO3. Determine the limiting reactant, theoretical yield, and percent yield for the reaction.

Molar masses, g/mol:

CaO = 56.08 ; CO2= 44.01 ; CaCO3= 100.09

Answer 1

Solution :

The given reaction equation is already balanced.

$\dfrac{15.27\;g}{56.08\;g}=0.272289\;moles\;of\;CaO$

$\dfrac{18.06\;g}{44.01\;g}=0.4103613\;moles\;of\;CO_2$

0.2722896 mole of CaO would react completely with 0.2722896 moles of CO2, but there is more CO₂ present than that, so CO₂ is in excess and CaO is the limiting reactant.

$(0.272289\;moles\;of\;CaO)*(\dfrac{1moles\;of\;CaCO_3}{1moles\;of\;CaO})*(100.0875g\;of\;CaCO_3)=27.25g\;of\;CaCO_3\;in\;theory$

$\dfrac{22.15g}{27.25g}=0.81276=81.28$ % yield CaCO₃