The correct option is D 90%
2C4H10+13O2→8CO2+10H2O
Number of moles of butane =5858=1 mol
As per stoichiometry,
2 mol of butane produces 8 mol of CO2
1 mol of butane will produce 82=4 mol of CO2
So, theoretical yield = 4 × molar mass of CO2
theoretical yield =4×44=176 g
% yield=Experimental yieldTheoretical yield×100
% yield=158.4176×100=90%