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1. Hi can any one help me on this problem???
What mass of 2AlO3 product will form if 15g of the metal is used?
2. (Original post by AmirahxRashid)
Hi can any one help me on this problem???
What mass of 2AlO3 product will form if 15g of the metal is used?

Hope this is correct.
Mol of Al= mass/Mr
Mol of Al = 15/27
Mol ratio of Al to Al2O3 is 1:1
So moles of Al = moles of Al2O3
So if Mass= moles x Mr
Then mass of Al2O3= (15/27) x (27+27+16+16+16)
Mass of Al2O3 = 56.66
Mass= 56.7g to 3s.f.
3. (Original post by Call_me_bo)
Hope this is correct.
Mol of Al= mass/Mr
Mol of Al = 15/27
Mol ratio of Al to Al2O3 is 1:1
So moles of Al = moles of Al2O3
So if Mass= moles x Mr
Then mass of Al2O3= (15/27) x (27+27+16+16+16)
Mass of Al2O3 = 56.66
Mass= 56.7g to 3s.f.
Thank you i got 63g so i knew that had to be wrong because the equaution goes 4Al +3O2 =2AlO3
4. (Original post by AmirahxRashid)
Thank you i got 63g so i knew that had to be wrong because the equaution goes 4Al +3O2 =2AlO3
WAIT!
Oh if that's the equation then the mol ratio is 2:1
So the answer is half of 56.66g
It's 28.3g

Sorry again. That is the answer.
Welcome
5. (Original post by Call_me_bo)
WAIT!
Oh if that's the equation then the mol ratio is 2:1
So the answer is half of 56.66g
It's 28.3g

Sorry again. That is the answer.
Welcome
Lol thank you ever soo much
I thought you had to add both of the reactants so ghat why i did it wrong. But wouldnt you have to add 15g to get the mass of the products at the end????
In so sorry for like disturbing you😯😯😯😯
Ps this is what i did so if its alright with u could u please tell me what i did wrong here???
6. (Original post by AmirahxRashid)

Lol thank you ever soo much
I thought you had to add both of the reactants so ghat why i did it wrong. But wouldnt you have to add 15g to get the mass of the products at the end????
In so sorry for like disturbing you😯😯😯😯
Ps this is what i did so if its alright with u could u please tell me what i did wrong here???
7. (Original post by Call_me_bo)
Thank you

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