Question

8. Butane is burned in a lighter at a rate of 0.24 mol/min in the following reaction.
2 C4H10(g) + 13/2 O2(g) --> 4 CO2(g)+ 5H20(g)

(a) What is the rate at which O2(g) is consumed and CO2(g) is produced?
(b) How long does 15.0 g of C4H10(g) take to burn?

Answer 1

The equation of the combustion of butane is not balanced. Corrected is
C4H10(g) + 13/2 O2(g) --> 4 CO2(g)+ 5H2O(g) or
2C4H10(g) + 13O2(g) --> 8CO2(g)+ 10H2O(g)
a. for O2 it is 0.2 mol/min C4H10 x (13 mols O2/2 mols C4H10) = ? min
for CO2 it is 0.2 mol/min C4H10 x (8 mols CO2/2 mols C4H10) = ? min

b. You have 15 g C4H10. mols C4H10 = 15/58 = about 0.3 but you need a better number.
0.2 mols/min C4H10 x # min = approx 0.3.
Solve for # min to burn the 15 g C4H10.