What mass of water is associated with the complete combustion of a #9.80*g# mass of butane?

1 Answer
Nov 6, 2016

Answer:

Approx. #15*g# of water are produced.

Explanation:

We need (i) a stoichiometrically balanced equation:

#C_4H_10(g) + 13/2O_2(g) rarr 4CO_2(g) + 5H_2O(l)#,

and (ii), the equivalent quantity of butane combusted:

#(9.80*g)/(58.12*g*mol^-1)# #=# #0.169*mol#.

The given equation CLEARLY shows that 5 equiv of water are procuced per equiv of butane.

Therefore, #"mass of water"#

#=# #0.169*molxx5xx18.01*g*mol^-1=15.2*g#

What mass of oxygen gas was required for this reaction?