What mass of water is associated with the complete combustion of a #9.80*g# mass of butane?

Question

Mateo Aguilar · Accepted Answer

Approx. #15*g# of water are produced.

(i) A stoichiometrically balanced equation is required. #C_4H_10(g) + 13/2O_2(g) rarr 4CO_2(g) + 5H_2O(l)#, and (ii), the corresponding amount of butane burned: #(9.80*g)/(58.12*g*mol^-1)# #=# #0.169*mol#. It is VERY CLEAR from the given equation that five equivalents of water are produced for every equivalent of butane. Therefore, #"mass of water"# #=# #0.169*molxx5xx18.01*g*mol^-1=15.2*g# How much oxygen gas in mass was needed for this reaction?

Jose Ayala · Answer

undefined The molar mass of butane is approximately 58.12 g/mol. The balanced chemical equation for the complete combustion of butane is:

2 C4H10 + 13 O2 → 8 CO2 + 10 H2O

Using stoichiometry, we can calculate the mass of water produced:

Moles of butane = mass of butane / molar mass of butane
Moles of butane = 9.80 g / 58.12 g/mol ≈ 0.1685 mol

According to the balanced equation, 2 moles of butane produce 10 moles of water.

Moles of water = 10/2 * moles of butane
Moles of water = 5 * 0.1685 mol = 0.8425 mol

Mass of water = moles of water * molar mass of water
Mass of water = 0.8425 mol * 18.015 g/mol ≈ 15.19 g

Therefore, approximately 15.19 grams of water is associated with the complete combustion of a 9.80 grams mass of butane.