To find the answers to these questions, we need to use the balanced equation provided and perform some calculations based on the given information.
Let's start with the first question:
1) How many kilocalories are released by the combustion of 17.7 g of C8H18?
To find the amount of kilocalories released, we need to determine the amount of octane consumed in the combustion reaction. Here's how you can do it:
Step 1: Calculate the moles of C8H18 using the molar mass.
Molar mass of C8H18 = (12.01 x 8) + (1.01 x 18) = 114.23 g/mol
Moles of C8H18 = mass / molar mass = 17.7 g / 114.23 g/mol = 0.155 mol
Step 2: Use the stoichiometric coefficients from the balanced equation to convert moles of C8H18 to moles of energy released.
From the balanced equation, we see that 2 moles of C8H18 produce 239.5 kcal.
So, 0.155 mol of C8H18 will produce (0.155 mol / 2 mol) * 239.5 kcal = 18.74 kcal.
Therefore, the combustion of 17.7 g of C8H18 would release approximately 18.74 kilocalories.
Moving on to the second question:
2) How many moles of octane must be burned to release 540.0 kcal?
To find the number of moles required, we can use the equation derived from the stoichiometric coefficients:
Moles of C8H18 = (kcal of energy required / kcal per 2 moles of C8H18) * (2 moles / kcal ratio)
From the balanced equation, we can see that 2 moles of C8H18 release 239.5 kcal. Therefore, the kcal ratio is 239.5 kcal / 2 moles.
Plugging in the values:
Moles of C8H18 = (540.0 kcal / 239.5 kcal) * (2 moles / 1 kcal) = 2.25 moles
Thus, to release 540.0 kcal, 2.25 moles of octane must be burned.
Now let's move on to the third question:
3) How much energy (in kJ) is released by the combustion of 1.77 mol of C8H18?
Using the stoichiometric coefficients from the balanced equation, we know that 2 moles of C8H18 release 1002 kJ of energy.
Therefore, the energy released by the combustion of 1.77 mol of C8H18 would be:
Energy = (1.77 mol / 2 mol) * 1002 kJ = 886.21 kJ
Hence, the combustion of 1.77 mol of C8H18 would release approximately 886.21 kJ of energy.