Al + Fe2O3 --> Al2O3 + Fe How many grams of Al2O3 would be produced when 10.0 grams of Al is reacted with an excess supply of Fe2O3? You must show work to get full credit.

18.88 grams of Al2O3

Explanation:

Here's my work

10g Al    l    1 mol Al    l    1 mol Al2O3  l   102g Al2O3

             l    27g Al       l   2 mol Al         l    1 mol Al2O3

Balanced equation: 2Al +Fe2O3  --> Al2O3 + 2Fe

So pretend it's in a chart,  you fill it out starting from the top left corner and then moving diagonally down and then straight up and then diagonally down and the straight up, etc.  You start with 10 grams of Al, so the first thing you want to do is convert it into moles.   So you find the mass of Al(which is 27) and then you have 1 mol of Al on top.  So the next thing you have to do is to change it to Al2O3 since that is what the question is asking for.  This is where it's important you balanced the equation correctly because you're going to compare the mol ratios.  To do this, you have to look at the coefficients in front of Al, which is a 2.  so you write 2 mol Al on the second bottom section.  and then on top you look at the balanced equation again and look at the coefficient in front of Al2O3, which there is none so it's a 1.   Then all you have to do is convert back to grams because the question wants it in grams.  so you put 1 mol on the bottom and then you look at your periodic table to figure out the mass of Al2O3.    Al= 27,  O= 16.   (27x 2) +  (16 x 3)  = 102.     Finally you just multiply all the numbers on the top row and then multiple all the numbers on the bottom row and then divide what you get on the top by the bottom.  

m=9

step-by-step explanation: