Given the information below, calculate the heat provided by combustion of 66.1g of propane (C3H8). ΔHf° (C3H8(g)) = -103.85 kJ ΔHf° (CO2(g)) = -393.5 kJ ΔHf° (H2O(ℓ)) = -285.8 kJ Can Someone explain this in detail? i have to answer this type of problems on my test.
C3H8 + 5O2 ==> 3CO2 + 4H2O
dHrxn = (n*dHf products) - (n*dHf reactants)
To calculate the heat provided by the combustion of propane (C3H8), we need to use the standard enthalpy of formation (ΔHf°) values for each substance involved in the reaction.
The balanced chemical equation for the combustion of propane is:
C3H8(g) + 5O2(g) -> 3CO2(g) + 4H2O(ℓ)
This equation tells us that one molecule of propane reacts with 5 molecules of oxygen gas to form 3 molecules of carbon dioxide gas and 4 molecules of liquid water.
Now, let's break down the steps to calculate the heat provided by the combustion:
Step 1: Calculate the molar mass of propane (C3H8).
Carbon (C) has a molar mass of 12.01 g/mol, and hydrogen (H) has a molar mass of 1.01 g/mol.
The molar mass of propane (C3H8) can be calculated as follows:
(3 * 12.01 g/mol) + (8 * 1.01 g/mol) = 44.11 g/mol
Step 2: Calculate the moles of propane combusted.
We were given the mass of propane as 66.1 g. To calculate the moles, divide the mass by the molar mass:
66.1 g / 44.11 g/mol = 1.5 mol
Step 3: Calculate the heat of combustion for propane.
The heat of combustion (ΔHc) can be calculated using the following equation:
ΔHc = Σ (ΔHf°products) - Σ (ΔHf°reactants)
First, we need to calculate the change in enthalpy for the products:
Σ (ΔHf°products) = (3 * ΔHf°CO2) + (4 * ΔHf°H2O)
Σ (ΔHf°products) = (3 * -393.5 kJ/mol) + (4 * -285.8 kJ/mol) = -2628.3 kJ/mol
Next, we calculate the change in enthalpy for the reactants:
Σ (ΔHf°reactants) = ΔHf°C3H8 + 5 * ΔHf°O2
Σ (ΔHf°reactants) = -103.85 kJ/mol + (5 * 0 kJ/mol) = -103.85 kJ/mol
Finally, substitute the values into the equation:
ΔHc = Σ (ΔHf°products) - Σ (ΔHf°reactants)
ΔHc = -2628.3 kJ/mol - (-103.85 kJ/mol) = -2524.45 kJ/mol
Step 4: Calculate the heat provided by the combustion of 66.1 g of propane.
To calculate the heat provided, multiply the heat of combustion (ΔHc) by the moles of propane combusted:
Heat provided = ΔHc * moles of propane combusted
Heat provided = -2524.45 kJ/mol * 1.5 mol = -3786.68 kJ
Therefore, the heat provided by the combustion of 66.1 g of propane is approximately -3786.68 kJ.
Remember to pay attention to the units and ensure they are consistent throughout the calculations.