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

What record? the question is incomplete