Answer:
12.6 g of H₂O can be produced in the combustion
Explanation:
This excersise involves an easy combustion reaction:
2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O
First step: calculate moles of reactants in order to find out the limiting.
23 g . 1mol / 58g = 0.396 moles of butane
29.1 g . 1mol /32g = 0.909 moles of O₂
2 moles of butane react to 13 moles of oxygen
Then, 0.396 moles of butane may react to (0.396 . 13) / 2 = 2.574 moles
Certainly we do not have enough oxygen, so O₂ becomes the limiting reactant
13 moles of O₂ can produce 10 moles of water
Then 0.909 moles may produce (0.909 . 10) /13 = 0.699 moles H₂O
We convert moles to mass → 0.699 mol . 18 g/mol = 12.6 g
