Question

The following equation shows the conversion of aluminum oxide (from the ore bauxite) to aluminum:
2Al2O3(s)→4Al(s)+3O2(g), ΔH = +801 kcal (+3350 kJ )

Part B:

How many kilocalories are required to produce 1.16 mol of aluminum?

Part C:

How many kilojoules are required to produce 1.16 mol of aluminum?

Part D:

How many kilocalories are required to produce 11.6 g of aluminum?

Part E:

How many kilojoules are required to produce 11.6 g of aluminum?

Answer 1

Here we just have to apply the stoichiometry rules, for our particular case we have been indicated that 801 kcal (3350 KJ) are consumed for the production of 4 aluminum moles,

For part B: the easiest way to visualize this is to find the energy to produce just 1 mole, to do this divide the energy absorbed by 4 so:

801 / 4 = 200.25 Kcal, this is the energy absorbed to produce 1 mole of aluminum, now we only need to multiply this value by 1.16 so:

200.25 * 1.16 = 232.29 Kcal this is the energy required to produce 1.16 moles of aluminum

For part C:

We repeat the procedure for kilojoules

3350 / 4 = 837.5 KJ / mol, this is the energy to produce 1 mole of aluminum, now multiply this by 1.16

971.5 KJ are required

You can also just multiply the value from part B of kcal and multiply it by 4.184 this is because 1 kcal = 4.184 KJ

For part D

we are asked about 11.6 grams, the easiest thing to do is to change this mass to moles by dividing this value by the molar mass (molar mass aluminum = 27 g/gmol)

moles = mass / molar mass

11.6 / 27 = 0.4296 moles of aluminum

in part B we got the value to produce 1 mole of aluminum which is 232.29 kcal, we only need to multiply

200.25 * 0.4296 = 86.03 Kcal for 11.6 grams of aluminum

Part E

We know that we require 86.03 Kcal for 11.6 grams of aluminum we also know that 1Kcal is equal to 4.184 KJ so

86.03 * 4.184 = 359.94 KJ for 11.6 grams of aluminum

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