In a butane lighter, 9.7 g of butane combine with with 34.7 g of oxygen to form g g carbon dioxide and 29.3 how many grams of water?

1 Answer
Jul 14, 2017

Answer:

Approx. #15*g#......

Explanation:

We need a stoichiometric equation........

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

And thus #"moles of butane"=(9.7*g)/(58.12*g*mol^-1)=0.167*mol#.

And so upon complete combustion we gets......

#0.167*molxx5xx18.011*g*mol^-1=15.0*g# water.......

We could have avoided this rigmarole by noting that mass is always with conserved in a chemical reaction. By the problem's specification, we had #(34.7+9.7)*g=44.4*g# of reactant; we got a mass #29.3*g# carbon dioxide, and thus there was a balance of #44.4-29.3*g=15.0*g# as required........WHICH MUST REPRESENT THE MASS OF WATER...........