If you have Al2O3 how many grams of aluminum oxide can be produced from 125g of aluminum. Can anyone please?

Question

tell me how to do this?thanks.

Answer 1

Ok so the equation would be balanced
4Al + 3 O2 = 2Al2O3
125g excess ?


T-Tables
125g Al (1Mol/ 27g al) (4 mol/1mol)= 18.5 mol of Al

we then convert it to moles of the products

18.5 mol Al (1mol Al/2mol Al) = 9.25 mol of AlO3 product
* we got the 1/2 ratio becasue
4 mol of Al (reactant)/ 2 mol Al (product)

9.25 mol Al (27 g Al/ 1 mol Al) = 250g AlO3

thats your answer hope i helped

Answer 2

Okay the way you do this is you need to compare apples with apples, so this means converting to moles of each product.

First find how many moles of Al you get with a 125g sample.
125g Al / [26.9g/mol] = 4.65 moles Al

Now find how many moles of Al2O3 this can make knowing that from the formula, it takes two Al moles to make a Al2O3.


4.65 mol Al / [2 mol Al / 1 mol Al2O3] = 2.33 mol Al2O3

Now convert this into a gram amount using the molar mass of Al2O3 which if you add the individual molar masses of 2 Al ions and 3 oxygen ions, you get 101.8g.

2.33 mol Al2O3 / [1 mol Al2O3 / 101.8g Al2O3] = 237.1g Al203

Hope that helps.

Answer 3

First, write a balanced equation:

4Al (s) + 3O2 (g) ----> 2Al2O3

Use factor label method:
Convert to moles. Multiply 125 g of Al by 1 mol over the molar mass of Al (26.98 g)
125g Al (1 mol Al / 26.98g Al )
4.633 mol Al

There are 2 moles of Al2O3 and 4 moles of Al, so they are in a 1/2 ratio.

Therefore, multiply 4.633 mol Al by (1 mol Al2O3 / 2 mol Al) to get
2.32 mol Al

Convert to grams. Use molar mass of Al2O3 = 101.96 g.

2.32 mol Al2O3 (101.96 g Al2O3 / 1 mol Al2O3)

236.1 grams

Answer 4

4Al(s) + 3O2(g) = 2Al2O3(s)

Moles(Al) = 125/ 27 = 4.63 (approx.) equivalent to '4'.
moles(Al2O3) = 4.63 x 2 / 4 = 2.31(approx) equivalent to '2'.

Mr(Al2O3) =
2 x Al = 2 x 27 = 54
3 x O = 3 x 16 = 48
54 + 48 = 102

Mass(Al2O3) = 2.31 x 102 = 236.1111.... g