Question

143 g of AL are reacted with 601 g of Fe2O3. Calculate the mass of Al2O3 formed and determine the amount of excess reagent left at the end of the reaction. 2Al + Fe2O3 ->Al2O3 + 2Fe

Answer 1

this question is pretty much the same as the one you asked just before.

Answer 2

determine the limiting reactant, and use the stoichiometric coefficients to build a ratio.

Answer 3

i didn't understand your explanation :/

Answer 4

do you know how to find the limiting reactant?

Answer 5

No, I think it is iron but the way I was taught was basically to just use which ever part was left out, which I know isn't the right way to be taught

Answer 6

you have to convert the mass given to moles (n)

\(n=\dfrac{m}{M}\) ; M= molar mass, m=mass, n=moles

then divide the moles by it's stoichiometric coefficient

normalized moles =\(\dfrac{mass}{coefficient}\)

Answer 7

I have no idea what any of that means

Answer 8

how is that the mass of Al2O3?

Answer 9

you have to work in moles to find the mass of Al2O3, it's not a one step question. Theres 5 or 6 steps to do.

Answer 10

i found the moles of Al and Fe2O3. what do i do after that?

Answer 11

find which one is the limiting reactant, because thats the amount of moles you'll be working with

Answer 12

okay. and what do i do with the amount of moles of the limiting reactant?