[Maverick]

so i have to find how much Al is used when there is 8 grams of Fe2O3 in this thermite reaction

1. Balance equation Fe₂O₃ + 2Al 2Fe + Al₂O₃

2. Define A, B X, Y
A = Fe2O3
B= Al
X = 1
Y= 2

3: Convert any given to moles 8g Fe2O3 / 160 g/mol = 0.05 mol

4 next multiply the moles from step 3 by (Y/X)
0.05 mol x (2/1) = 0.033 moles of Al

5. convert to grams
0.033 moles x 26.982 = 0.89 g used

but the receipe for thermite calls for a ratio of 8:3 of iron oxide:Al, why didnt i get 3 grams as the amount of Al used im confused a bit. plz correct if i did anything wrong

[rabolisk]

0.05 mol x (2/1) = 0.033 moles of Al

Error.

[Maverick]

0.05 mol x (2/1) = 0.033 moles of Al

Error.

0.05moles Fe2O3 x (2Al / 1Fe2O3) = .1 moles of Al used

so .1 mol x 27 g/mol = 2.7 grams. thank you for pointing that out

[Maverick]

i feel stupid now, gotta be careful with those calculations. /carelessness