Question

C3Hg(g) + 5O2(g)-> 3CO2(g)+ 4H2O(g) delta H= -2044 kJ

how many grams of propane must be reacted to release 7563 kJ of heat

Answer 1

To find out how many grams of propane must be reacted to release 7563 kJ of heat, we use the combustion reaction's enthalpy change and molar mass of propane to calculate the required amount through stoichiometry.

Step-by-step explanation:

The student is asking how many grams of propane, C3H8, must be reacted to release 7563 kJ of heat. Given the enthalpy change (ΔH) of the combustion reaction, which is -2044 kJ, we can calculate this using stoichiometry and proportionality.

Firstly, we establish the relationship between the amount of heat released and the moles of propane. We know that the combustion of 1 mole of propane releases 2044 kJ of heat:

1 mole C3H8 → -2044 kJ

To find out how many moles are needed to release 7563 kJ, we set up the proportion:

x moles C3H8 → -7563 kJ

Solving for x gives us the number of moles of propane required. Then, we convert the moles of propane to grams using the molar mass of propane, which is 44.1 g/mol:

Number of grams = Moles of propane × Molar mass of propane

Finally, we arrive at the required mass of propane needed to release 7563 kJ of heat.