How much heat energy is produced when 31.6 grams of propane are burned?
I asked my teacher and she said it would be a stochiometry problem but im still confuded:/
*confused lol(:
Yes. I wondered if confuded was a new word I should learn.
I think the problem is a stoichiometry problem IF you know the heat of combustion. Do you know that? If so, then
C3H8 + 5O2 ==> 3CO2 + 4H2O
So it takes 44 g C3H8 to generate ??kJ (the heat of combustion generally given in kJ/mol), then
heat produced by 44 g = ?? kJ x (31.6/44) =
To solve this problem, you are correct that it involves stoichiometry, which is the calculation of quantitative relationships between reactants and products in a chemical reaction. In this case, propane (C3H8) is being burned, which means it reacts with oxygen (O2) to form carbon dioxide (CO2) and water (H2O).
To determine the amount of heat energy produced when 31.6 grams of propane are burned, follow these steps:
1. Write the balanced chemical equation for the combustion of propane:
C3H8 + 5O2 => 3CO2 + 4H2O
2. Calculate the molar mass of propane (C3H8):
C = 12.01 g/mol (carbon) x 3 = 36.03 g/mol
H = 1.01 g/mol (hydrogen) x 8 = 8.08 g/mol
Total molar mass of propane = 36.03 g/mol + 8.08 g/mol = 44.11 g/mol
3. Convert grams of propane to moles:
Moles of propane = 31.6 g / 44.11 g/mol = 0.7164 mol (rounded to four decimal places)
4. Use the stoichiometry ratios from the balanced equation to calculate the moles of water produced:
From the balanced equation, we see that 1 mole of propane produces 4 moles of water.
Moles of water produced = 0.7164 mol propane x (4 mol water / 1 mol propane) = 2.8656 mol (rounded to four decimal places)
5. Convert moles of water produced to grams:
Grams of water = moles of water x molar mass of water
Molar mass of water = 18.015 g/mol (2 hydrogen atoms at 1.01 g/mol + 1 oxygen atom at 16.00 g/mol)
Grams of water = 2.8656 mol x 18.015 g/mol = 51.761 g (rounded to three decimal places)
6. Calculate the heat energy produced using the heat of combustion:
The heat of combustion of propane is typically -2220 kJ/mol (negative sign indicates heat is released).
Heat energy produced = moles of propane x heat of combustion
Heat energy produced = 0.7164 mol x (-2220 kJ/mol) = -1580 kJ (rounded to the nearest whole number)
Therefore, when 31.6 grams of propane are burned, approximately -1580 kJ of heat energy is produced. The negative sign indicates that the reaction releases heat energy.