HOW MANY GRAMS OF OCTANE WOULD BE NEEDED TO RELEASE 3,500 kJ OF HEAT?

2C8H18 + 25O2-----> 16CO2 + 18H2O

DeltaH= -5,530 kJ/mole of reaction

one mole gives you 5530kj

so you want then 3500/5530 of one mole
mass then= 3500/5530 * molemass Octane

To find out how many grams of octane are needed to release 3,500 kJ of heat, we need to use the balanced chemical equation and the given value of ΔH. Here are the steps to calculate it:

1. Find the molar quantity of octane (C8H18) in the balanced equation.
In the balanced equation, the coefficient of octane (C8H18) is 2, which means that 2 moles of octane are needed for the reaction.

2. Convert the given heat energy from kJ to moles using the balanced chemical equation.
We know that the ΔH for the reaction is -5,530 kJ per mole of reaction. So, we can set up the following ratio:
-5,530 kJ ΔH / 1 mole of reaction = 3,500 kJ / x moles of reaction (where x is the unknown quantity we're trying to find)

Cross-multiplying, we get:
-5,530 kJ × x = 3,500 kJ × 1
x = (3,500 kJ × 1) / -5,530 kJ
x ≈ -0.632 moles

Since we can't have a negative quantity in this context, it suggests that the reaction is exothermic. Thus, 0.632 moles of octane are needed to release 3,500 kJ of heat.

3. Calculate the grams of octane required using the molar mass of octane (C8H18).
The molar mass of octane can be calculated by summing up the atomic masses of carbon and hydrogen in one mole of octane:
(12.01 g/mol × 8) + (1.01 g/mol × 18) = 114.23 g/mol

Finally, we can convert the moles of octane to grams using the molar mass:
0.632 moles × 114.23 g/mol ≈ 72 grams

Therefore, approximately 72 grams of octane would be needed to release 3,500 kJ of heat.