In the reaction shown above, 2.00 x 10^3 g caco3 produce 1.05 x 10^3 g of cao, what is the percent yield?

First, we have to get the theoretical yield of CaO:
the balanced equation for the reaction is:
CaCO3(s)→CaO(s) +CO2(g)
covert mass to moles:
moles CaCO3 = mass of CaCO3 / molar mass of CaCO3
                         = 2x10^3 /100 = 20 moles
the molar ratio between CaCO3 : CaO = 1:1
∴moles of CaO = 1* 20 = 20 moles 
∴mass of CaO = moles of CaO * molar mass of CaO
                        = 20 * 56 = 1120 g
∴the theoritical yield = 1120 g and we have the actual yield =1.05X10^3
∴Percent yield = actual yield / theoritical yield *100
                         = (1.05x10^3) / 1120 *  100 
                         = 94 %
 

Most likely a is the answer - because  you generate   38 atp's but also consume an atp.   each atp gives approx. less than 4 kcal so probably a is the closest answer.