Quicklime (CaO) can be prepared by roasting limestone (CaCO3)?

Question

according to the reaction  CaCO3(s)---> CaO(s) + CO2(g)
When 2.7 ×103 g of CaCO3 are heated, the
actual yield of CaO is 1 ×103 g. What is the
percent yield?
Answer in units of %

John · Accepted Answer

Inspection of the balanced equation shows that 1 mole of CaCO3 produces 1 mole CaO. Stoichiometry yields:  
2.7 x 10^3 g CaCo3  x  1 mole CaCo3    x    1 mole CaO     x 56.1 g CaO      = 1.0 x 10^ 3 g CaO
_______________________________________________________________
                                 100.1 g CaCO3   x  1 mole CaCO3   x 1 mole CaO

Since the actual yield was 1 x 10^3 g, % yield is 1.0/1.5  x 100 = 67%

1. Use stoichiometry to determine the quantity that should have been produced 
2. then divide quantity produced by theoretical yield, multiply by 100 to convert fraction to a percent

Simple and straight forward, study this solution, then try some on your own.

cat lover · Answer

Based on the balanced reaction one mole of calcium carbonate will give one mole of calcium oxide.

So how many moles of calcium carbonate do you have?

How many moles of calcium oxide does 1 x 10^3 grams equal?

So if you had 100 moles of calcium carbonate, and got 95 moles of calcium oxide as product, then the percent yield is 95%  (And no, I didn't do the arithmetic, so don't depend on my numbers.)

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Quicklime (CaO) can be prepared by roasting limestone (CaCO3)?

2 Answers