Answer to Question #173366 in General Chemistry for RED badilla

Question #173366

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


1
Expert's answer
2021-03-22T06:34:00-0400

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



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