Question

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

Answer 1

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

Answer 2

2Al is the limiting reagent right

Answer 3

did you check to see?

Answer 4

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

Answer 5

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

Answer 6

or limiting aent

Answer 7

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

Answer 8

alright and then there is 2 aluminum molecules per 2 iron

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

Al:Fe 1:1

Answer 13

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

Answer 14

yes

Answer 15

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

Answer 16

now just change 5.7 moles of Fe to grams

Answer 17

m=5.7*55=322g

Answer 18

so that would be your answer

Answer 19

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

Answer 20

answer not correct i think i made a mistake

Answer 21

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

Answer 22

174.85 s the answer thanks