How do you determine the mass of carbon dioxide produced when 0.85 g of butane reacts with oxygen according to the following equation: #2C_4H_10 + 13O_2 -> 8CO_2 10H_2O#?

1 Answer
Sep 6, 2016

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

#5.58xx10^-2 "moles of "CO_2" are produced".#

Explanation:

#"Moles of butane"# #=# #(0.85*g)/(58.12*g*mol^-1)# #=# #1.46xx10^-2mol#.

Given the stoichiometry, each moles of butane gives #4*mol# #CO_2# upon complete combustion.

And thus #"moles of "CO_2# #=# #4xx1.46xx10^-2mol=5.58xx10^-2mol#.

This is an equivalent mass of #5.58xx10^-2molxx44.01*g*mol^-1~=2.5*g#.

In my stoichiometric equation, I used an half equiv of dioxygen in order to balance the equation. I do this because it makes the 'rithmetic a bit easier.