## Jonask 2 years ago Calculate the mass of iron metal (in grams) that can be prepared from 150 grams of aluminum and 250 grams of iron(III) oxide.

\[ {F}e_2O_3+2Al\implies Al_2O_3+2Fe\]

2Al is the limiting reagent right

3. Outkast3r09

did you check to see?

novice in chem,well i just looked at the atom it has 2 moles and Fe2O3 has 1 so yeah

5. Outkast3r09

to figure out which is limiting you want to to take Grams of reactant ->mols of reactant->moles of product with both... which ever can make the least amount of moles of product is your limiting reactant

6. Outkast3r09

or limiting aent

Fe2O3 M=55 m=250 n=4.5 Al M=26 m=150 n=5.7

8. Outkast3r09

alright and then there is 2 aluminum molecules per 2 iron

9. Outkast3r09

so you cna make 9.16 moles of iron with 9.16 moles of aluminum

wait 9.6 was error i took wrong mass Al n=5.7

11. Outkast3r09

o wait nvm 5.7 molecules of aluminum can make 5.7 molecules for iron

Al:Fe 1:1

13. Outkast3r09

however per each iron oxide molecule you can make 2 iron oxide molecules

yes

15. Outkast3r09

so multiply 4.5*2 =9 moles of iron.... since you can only make 5.7 moles with the aluminum , Al is your limiting reagent

16. Outkast3r09

now just change 5.7 moles of Fe to grams

m=5.7*55=322g

18. Outkast3r09

so cant we say there 2Al n=5.7 but 2 moles n=2(5.7)

the periodic table i am using says atomic weight =Al M=26g/mol is this the right thing to use as M