Chemistry honors:
calculations involving a limiting reactant
for each of the following unbalanced chemical equations, suppose that exactly 5.00 g of each reactant is taken. determine which reactant is limiting, and calculate what mass of each product is expected (assuming that the limiting reactant is completely consumed).
s(s) + 2h2so4(aq) --> 3so2(g) + 2h2o(l)

In order to determine the limiting reactant, we need to obtain the number of moles of each reactant and dividing this value with the stoichiometric coefficient of the reactant. The lowest value would pertain to the limiting reactant. 

For S (32 g/mol)

number of moles of S = 5/32 = 0.15625
divide by the stoichiometric coefficient,1 = 0.15625

For H2SO4 

number of moles of H2SO4 = 0.0510
divide by the stoichiometric coefficient,2 = 0.02551

Therefore the limiting reactant is H2SO4

Mass of each product when limiting reactant is fully consumed

mass of SO2 = 0.0510 mol H2SO4 * (3 mol SO2/ 2 mol H2SO4) * (64 g SO2/ 1mol SO2) = 4.896 g

mass of H2O = 0.0510 mol H2SO4 * (2 mol H2O/ 2 mol H2SO4) * (18 g H2O/ 1 mol H2O) = 0.918 g H2O

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.