Answer to Question #161982 in General Chemistry for Kate

Question

if 55.0 grams of aluminum powder is mixed with 75.0 grams of ferric acid, which of the two substance is the limiting reagent? How many grams of molten iron will be produced?

Answer 1

Solution:

Ferric acid does not exist. And if it did exist its reaction with aluminum would not produce molten iron. So supposing this question is really about reacting ferric oxide with aluminum, which is a useful reaction that produces molten iron:

given chemical reaction:

2Al + Fe₂O₃ → Al₂O₃ + 2Fe

$\dfrac{(55.0 g Al)}{27\;(\frac{g}{mole})}=2.03843\;moles\;of\;Al$

$\dfrac{(75.0 g \;ofFe_2O_3)}{160\;(\frac{g}{mole})}=0.469665\;moles\;of\;Fe_2O_3$

0.469665 mole of Fe2O3 would react completely with 0.469665 x ($\frac{2}{1}$) = 0.93933 mole of Al, but there is more Al present than that, so Al is in excess and Fe₂O₃ is the limiting reagent.

m=(0.469665 mol Fe2O3) x ($\tfrac{2\;moles\;Fe}{1\;mole\;Fe_2O_3}$) x (55.8450 $(\frac{g}{mole})$ ) = 52.5 g of Fe