To solve this problem, we need to use the ideal gas law to determine the number of moles of butane (C4H10) that are present in the given volume and conditions. Then, we can use the balanced chemical equation to find the ratio of moles of C4H10 to moles of H2O produced, and finally use the molar mass of water to convert the number of moles to grams.
Step 1: Calculate the number of moles of butane
Using the ideal gas law:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.
We need to convert the given temperature from Celsius to Kelvin by adding 273.15:
T = 65.0°C + 273.15 = 338.15 K
Now we can solve for n:
n = PV/RT = (1.70 atm)(36.0 L)/(0.0821 L·atm/mol·K)(338.15 K) = 1.56 mol
So we have 1.56 moles of C4H10.
Step 2: Calculate the number of moles of water produced
From the balanced chemical equation:
2C4H10 + 13O2 → 8CO2 + 10H2O
we see that 2 moles of C4H10 produce 10 moles of H2O. Therefore, the ratio of moles of C4H10 to moles of H2O is:
2 mol C4H10 / 10 mol H2O = 0.2 mol C4H10 per mol H2O
So for 1.56 moles of C4H10, we will have:
1.56 mol C4H10 × (1 mol H2O / 0.2 mol C4H10) = 7.8 mol H2O
Step 3: Convert moles of H2O to grams
Using the molar mass of water, which is 18.015 g/mol, we can convert the number of moles of H2O to grams:
7.8 mol H2O × 18.015 g/mol = 140.3 g H2O
Therefore, 140.3 grams of water will be produced.