Step-by-step explanation:
The balanced chemical equation for the combustion of butane is:
2C4H10 + 13O2 → 8CO2 + 10H2O
From the equation, we can see that for every 2 moles of butane that react, 10 moles of water are produced.
To find the mass of water produced when 5.13g of butane reacts, we need to first convert the mass of butane to moles:
5.13g of butane / molar mass of butane (58.12 g/mol) = 0.0883 moles of butane
Since there is excess oxygen, we can assume that all of the butane will react completely, so we can use the mole ratio from the balanced equation to find the number of moles of water produced:
0.0883 moles of butane × (10 moles of water / 2 moles of butane) = 0.4415 moles of water
Finally, we can convert the moles of water to grams:
0.4415 moles of water × molar mass of water (18.02 g/mol) = 7.95g of water
Therefore, the mass of water produced when 5.13g of butane reacts with excess oxygen is 7.95g.
Hopes this helps