Final answer:
The mass of water produced when 3.53 g of butane combusts with excess oxygen is calculated to be 5.47 g, using stoichiometry and the balanced chemical equation for the combustion of butane.
Step-by-step explanation:
Calculation of Mass of Water Produced
To calculate the mass of water produced when 3.53 g of butane (C4H10) reacts with excess oxygen (O2), we need to use the balanced chemical equation for the combustion of butane:
C4H10 + 6.5O2 → 4CO2 + 5H2O
First, we find the molar mass of butane: (4 × 12.01) + (10 × 1.008) = 58.14 g/mol.
We then calculate the moles of butane used:
Moles of C4H10 = mass / molar mass = 3.53 g / 58.14 g/mol = 0.0607 mol
From the balanced equation, the mole ratio of C4H10 to H2O is 1:5. Therefore, moles of H2O produced = 5 × moles of C4H10 = 5 × 0.0607 mol = 0.3035 mol.
Finally, the mass of water produced is calculated using the molar mass of water: 18.015 g/mol.
Mass of H2O = moles × molar mass = 0.3035 mol × 18.015 g/mol = 5.47 g (rounded to three significant figures).