A student combusts 95.70g of butane (#C_4H_10#) in the presence of 94.40g of oxygen. How do you write and balance the equation for the reaction?

1 Answer
anor277
Feb 24, 2016

For complete combustion:

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

Explanation:

#"Moles of butane"# #=# #(95.7*g)/(58.12*g*mol^-1)# #=# #??# #"mol"#.

#"Moles of oxygen"# #=# #(94.4*g)/(32.00*g*mol^-1)# #=# #??# #"mol"#.

It is pretty clear that there is insufficient oxygen for complete combustion to occur. Butane would combust incompletely in these circumstances to give elemental carbon (as soot) and carbon monoxide gas. We have no handle on the amounts of each.

Related questions
See all questions in Balancing Chemical Equations
Impact of this question
1357 views around the world
You can reuse this answer
Creative Commons License