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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

jjennylove Feb 15, 2019

#1**+1 **

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

.ChowMein Feb 16, 2019

#2**+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