Final answer:
The maximum mass of water that could be produced from the reaction of 40.1 g of butane with 89.0 g of oxygen is 38.5 grams, based on the stoichiometry of the combustion reaction. Oxygen is the limiting reactant in this scenario.
Step-by-step explanation:
To determine the maximum mass of water that could be produced from the reaction of butane with oxygen, we first need to write the balanced chemical equation for the combustion of butane (C4H10):
2 C4H10(g) + 13 O2(g) → 8 CO2(g) + 10 H2O(g)
From the equation, 2 moles of butane react with 13 moles of oxygen to produce 10 moles of water. The molar mass of butane is 58.12 g/mol. Therefore, 40.1 g of butane is:
(40.1 g C4H10) / (58.12 g/mol) = 0.69 moles of C4H10
The molar mass of oxygen is 32.00 g/mol. Thus, 89 g of oxygen is:
(89 g O2) / (32.00 g/mol) = 2.78 moles of O2
Using the stoichiometry of the reaction, we calculate the moles of water produced from each reactant:
From butane: (0.69 moles C4H10) x (10 moles H2O / 2 moles C4H10) = 3.45 moles of H2O
From oxygen: (2.78 moles O2) x (10 moles H2O / 13 moles O2) = 2.14 moles of H2O
The limiting reactant is the one that produces the least amount of product, which in this case is oxygen. Therefore, the maximum amount of water produced is based on the moles of oxygen available:
(2.14 moles H2O) x (18.015 g/mol) = 38.5 g of H2O
Thus, the maximum mass of water that can be produced is 38.5 grams.