Right option is (c) 42.1%
To explain: The reaction is C5H12 + 8O2 -> 5CO2 + 6H2O. Basis: 10 moles of C5H12, => moles of O2 reacted = 80, => moles of O2 entered the process = 160, => moles of CO2, H2O and O2 in products are 50, 60 and 80, => percentage of H2O = 80/190*100 = 42.1%.