Answer: 3.54 g of water
Explanation: When 2.28 g of butane reacts with excess oxygen, the mass of water produced can be calculated using the balanced chemical equation for the combustion of butane.
The balanced equation for the combustion of butane is:
2 C4H10 + 13 O2 -> 8 CO2 + 10 H2O
From the equation, we can see that for every 2 moles of butane (C4H10), we get 10 moles of water (H2O).
To find the mass of water produced, we need to determine the number of moles of butane in 2.28 g.
The molar mass of butane (C4H10) is 58.12 g/mol.
To find the number of moles of butane, we can use the formula:
Number of moles = Mass / Molar mass
Number of moles of butane = 2.28 g / 58.12 g/mol
Number of moles of butane ≈ 0.0393 mol
Since the molar ratio of butane to water is 2:10, we can calculate the number of moles of water produced using the ratio:
Number of moles of water = Number of moles of butane x (10/2)
Number of moles of water ≈ 0.0393 mol x 5
Number of moles of water ≈ 0.1965 mol
Finally, we can calculate the mass of water produced using the formula:
Mass = Number of moles x Molar mass
Mass of water = 0.1965 mol x 18.02 g/mol
Mass of water ≈ 3.54 g
Therefore, when 2.28 g of butane reacts with excess oxygen, approximately 3.54 g of water is produced.