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

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

Chemistry
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\[ {F}e_2O_3+2Al\implies Al_2O_3+2Fe\]
2Al is the limiting reagent right
did you check to see?

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Other answers:

novice in chem,well i just looked at the atom it has 2 moles and Fe2O3 has 1 so yeah
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
or limiting aent
Fe2O3 M=55 m=250 n=4.5 Al M=26 m=150 n=5.7
alright and then there is 2 aluminum molecules per 2 iron
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
o wait nvm 5.7 molecules of aluminum can make 5.7 molecules for iron
Al:Fe 1:1
however per each iron oxide molecule you can make 2 iron oxide molecules
yes
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
now just change 5.7 moles of Fe to grams
m=5.7*55=322g
so that would be your answer
so cant we say there 2Al n=5.7 but 2 moles n=2(5.7)
answer not correct i think i made a mistake
the periodic table i am using says atomic weight =Al M=26g/mol is this the right thing to use as M
174.85 s the answer thanks

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