The combustion reaction of c3h8o with o2 is represented by this balanced chemical equation. 2c3h8o + 9o2 6co2 + 8h2o when 3.00 g of c3h8o and 7.38 g of o2 are combined, how many moles of which reactant remain? (a) 0.006 mol o2 (b) 0.024 mol c3h8o (c) 0.24 mol o2 (d) 0.18 mol c3h8o

According to the question - 1 mol C3H8O = 60.096 g C3H8O 2 mol C3H8O = 9 mol O2 1 mol O2 = 31.998 g O2 [ (3.00 g C3H8O) / 1][ (1 mol C3H8O) / (60.096) ][ (9 mol O2) / (2 mol C3H8O) ][ (32.998 g O2) / (1 mol O2) ] = 7.1880435 g O2 Since 7.1880435 g of O2 is needed, and 7.38 g of O2 is available, 0.199565 g of O2 will be left over and oxygen is present in excess. Next, we need to convert 0.199565 g of O2 into moles of O2: [ (0.199565 g O2) / 1][ (1 mol O2) / (31.998 g O2) ] = 0.005999 mol O2, or 0.006 mol O2