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In the reaction shown below , 2.00 x 10^3 g CaCO3 produce 1.05 x 10^3 g of CaO, what is the percent yield?

CaCO3 → CaO + CO2

 Feb 15, 2019
 #1
avatar+114 
+1

\(\text{Find theoretical yield}\\ 2*10^3g\;CaCO_3*\frac{1mol\;CaCO_3}{100.09g\;CaCO_3}*\frac{1mol\;CaO}{1mol\;CaCO_3}*\frac{56.08g\;CaO}{1mol\;CaO}=1120.59g\;CaO\\ Percent\;Yield=\frac{Actual\;Yield}{Theoretical\;Yield}*100\%=\frac{1.05*10^3g\;CaO}{1120.59g\;CaO}*100\%=93.7\%\)

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 Feb 16, 2019
 #2
avatar+708 
+1

For part 1 ) in the mole equation you have wouldnt it be 1.1 x10^3 answer since you multiply on top and divide at the bottom ?

jjennylove  Feb 16, 2019
 #3
avatar+114 
0


the intermediate calculations shouldn't be rounded, i just ended it at 1120.59 because i don't want to make it too long, round at the end.

the answer is limited to 3 sig-figs, so the final solution is rounded to 3 sig-figs, not the intermediate calculations

ChowMein  Feb 16, 2019
 #4
avatar+708 
0

i keep getting the value 1.11159564 , rounded it would 1.11

jjennylove  Feb 16, 2019
 #5
avatar+708 
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2 * 56.08=112.16/100.9 = 1.11159564

jjennylove  Feb 16, 2019
 #6
avatar+114 
0

Try multiplying by 2e3 instead of 2

ChowMein  Feb 16, 2019
 #7
avatar+708 
+1

I got it thank you! I understand all this part now smiley

jjennylove  Feb 16, 2019

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