How many grams of aluminum oxide is formed when 6.4 grams of aluminum is reacted?

Answer 1

To answer this first of all you need to know the equation of the reaction.

4Al + 3O2 ----> 2Al2O3

Then you need to know how many moles of aluminium you have mass = moles x molecular weight, therefore: moles = mass/ molecular weight. 38.8/21 = 1.85 (with rounding). This is the number of moles of aluminium you have.

To work out how much oxygen is added you need to take the ratio of moles(O2:Al) from the equation (3/4 or 0.75) and multiply by the number of moles of aluminium 1.85 x 0.75 = (5.55/4). This is the number of moles of oxygen. I left it like this as it will be nicer to multiply out later.

As you want the final mass you need to find the mass of oxygen added to the aluminium using mass = moles x molecular weight from before except as there are two oxygen atoms per molecule so the number needs to be doubled. (5.55/4) x (16 x 2) = 44.4g. This how much oxygen is added in the creation of aluminium oxide.

To get the final total you add the amount of aluminium to the amount of oxygen added 38.8g + 44.4g = 83.2g

Answer 2

(6.4 g Al) (moles Al/26.98 g Al) (1 mole Al2O3/2 moles Al) (101.96 g Al2O3/moles Al2O3) = 12. 09 g Al2O3 or 12 g (significant figures)

Steps:

1. Change everything to moles: use molar mass/molecular weight of Al

2. Find the molar ratio: since there are 2 Al in every Al2O3, then the ratio of Al2O3 to Al is 1:2

3. Change moles back to grams: use the molar mass/molecular weight of Al2O3

4. Watch out for significant figures!

Answer 3

The theoretical yield of aluminum oxide is 1.90 mol if 3.80 mol of aluminum metal is exposed to 3.15 mol of oxygen.

Answer 4

5.2 g of AL2

Answer 5

1.40 moles

Answer 6

5

Answer 7

7.20g

Answer 8

2.87g

Answer 9

0.185