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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?

Answer 1

For complete combustion:

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

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

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

We can see that there is not enough oxygen present for complete combustion to take place because butane would only partially burn under these conditions, producing elemental carbon (as soot) and carbon monoxide gas, both of which we cannot control in quantity.

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Answer 2

Caleb Adams

2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(g)

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.