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.
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
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???
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???