Jonask
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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Jonask
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\[ {F}e_2O_3+2Al\implies Al_2O_3+2Fe\]
Jonask
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2Al is the limiting reagent right
Outkast3r09
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did you check to see?
Jonask
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novice in chem,well i just looked at the atom it has 2 moles and Fe2O3 has 1 so yeah
Outkast3r09
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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
Outkast3r09
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or limiting aent
Jonask
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Fe2O3
M=55
m=250 n=4.5
Al M=26 m=150
n=5.7
Outkast3r09
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alright and then there is 2 aluminum molecules per 2 iron
Outkast3r09
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so you cna make 9.16 moles of iron with 9.16 moles of aluminum
Jonask
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wait 9.6 was error i took wrong mass Al n=5.7
Outkast3r09
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o wait nvm 5.7 molecules of aluminum can make 5.7 molecules for iron
Jonask
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Al:Fe 1:1
Outkast3r09
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however per each iron oxide molecule you can make 2 iron oxide molecules
Jonask
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yes
Outkast3r09
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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
Outkast3r09
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now just change 5.7 moles of Fe to grams
Jonask
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m=5.7*55=322g
Outkast3r09
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so that would be your answer
Jonask
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so cant we say there 2Al
n=5.7
but 2 moles
n=2(5.7)
Jonask
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answer not correct i think i made a mistake
Jonask
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the periodic table i am using says atomic weight =Al
M=26g/mol is this the right thing to use as M
Jonask
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174.85 s the answer thanks