What is the mass of water produced when 9.33 g of butane reacts with excess oxygen?

1 Answer
Mar 21, 2016

Answer:

#5xx0.161*molxx18.01*g*mol^-1=??#

Explanation:

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

#"Moles of butane"# #=# #(9.33*g)/(58.12*g*mol^-1)# #=# #0.161*mol#.

We assume complete combustion, and know that each mole of butane gives off 5 mol of water upon combustion.

Given that we know the molar quantity of butane, we simply perform the operation, #5xx0.161*molxx18.01*g*mol^-1=??#

What is the mass of carbon dioxide gas evolved in the reaction?