Calculate the mass of water produced when 4.86 g of butane reacts with excess oxygen.

The mass of water produced when 4.86 g of butane (C4H10) react with excess oxygen is calculated as below

calculate the moles of C4H10 used = mass/molar mass

moles = 4.86g/58 g/mol = 0.0838 moles

write a balanced equation for reaction

2 C4H10 + 13 O2 = 8 CO2 + 10 H2O

by use of mole ratio between C4H10 to H2O which is 2:10 the moles of

H20 = 0.0838 x10/2 = 0.419 moles of H2O

mass = moles x molar mass

=0.419 molx 18 g/mol = 7.542 grams of water is formed