C3H8 + 5O2 --> 3C02 + 4H2O contunite my question in add details?

Question

calculate the gram of excess reacgent units remaining when 150 gram of O2 molecules and 35.3 gram of C3H8 molecules react?

Answer 1

Find which is in excess (knowing that one mole of propane reacts with 5 moles of oxygen):

35.3 grams/58.12 g/mole = 0.607 moles of C3H8
0.607 moles X 5 = 3.035 moles O2 needed
150 g / 16.00 g/mole = 9.375 moles O2 present.
9.375 moles - 3.035 moles used = 6.34 moles O2 in excess
6.34 moles X 16.00 g/mole = 101.44 grams O2 in excess.