Final answer:
In the complete combustion of propane, the balanced equation is C3H8 (g) + 5O2(g) → 3CO2(g) + 4H2O(l). To completely combust 25.0 grams of propane, it would require a volume of air of 10.31 L at 25 °C and 1.00 atm. The heat of formation of propane can be calculated using the heat of combustion of propane, which is -1,131.9 kJ/mol. The increase in temperature of the water can be calculated using the equation q = m•C•ΔT.
Step-by-step explanation:
(a) The balanced equation for the complete combustion of propane gas is: C3H8 (g) + 5O2(g) → 3CO2(g) + 4H2O(l)
(b) To calculate the volume of air needed to completely combust 25.0 grams of propane, we first need to determine the moles of propane.
Using the molar mass of propane (44.1 g/mol), we find that there are 0.567 mol of propane. Since propane combusts with 5 moles of O2, we need 2.836 mol of O2. Using the information that 1.00 L of air at 25 °C and 1.00 atm contains 0.275 g of O2 per liter, we can calculate the volume of air needed as 10.31 L.
(c) The heat of combustion of propane is -2,219.2 kJ/mol. To calculate the heat of formation of propane, we can use the equation:
ΔHf(propane) = ΔHc(propane) - [ΔHf(H2O(I)) + 3ΔHf(CO2(g))]. Substituting the known values:
ΔHf(propane) = -2,219.2 kJ/mol - [(-285.8 kJ/mol) + 3(-393.5 kJ/mol)] = -2,219.2 kJ/mol + 1,087.3 kJ/mol = -1,131.9 kJ/mol.
(d) To calculate the increase in temperature of the water, we can use the equation:
q = m•C•ΔT, where q is the heat transferred, m is the mass of water, C is the specific heat capacity of water, and ΔT is the change in temperature.
The heat transferred can be calculated using the heat of combustion of propane (which is the heat released) and the moles of propane burned. The mass of water is given as 4.00 kilograms. The specific heat capacity of water is 4.18 J/g°C. Plugging in these values, we can solve for ΔT.