Calculate the mass of water produced when 1.92 g of butane reacts with excess oxygen.

C4h10+6.5o2=4co2+5h2o

moles of butane=1.92/58=0.0331 moles

moles of water=0.1655 moles/

as the butane and water has 1 is to 5 molar ratio

0.1655=mass/18

mass=2.98 g

mass of water produced = 2.98 g