Using : Fe2O3 + 2Al ---> 2Fe + Al2O3
How many grams of Al are needed to completly react with 135g Fe2O3?
How do i do this, i have no clue!
2007-02-05 11:57:15 · 4 answers · asked by Eddie G 1 in Science & Mathematics ➔ Chemistry
ok... what you have with the balanced equation esentially is the mole ratio of reactants and products.
you have 1 mol of Fe2O3 reacting with 2 moles of Al
now you have to convert g of Fe2O3 to moles and then x 2 to get the moles of Al then conver that to grams(and also find the atomic mass of Fe2O3)
135gFe2O3 x 1mol Fe2O3/ 159.7gFe2O3 x 2mols Al/ 1mol Fe2O3 x 26.98gAl/ 1mol Al = 45.61g of aluminum
hey alexis there are 2 moles of Al reacting with the 1 mol Fe2O3 not the other way around
2007-02-05 12:18:23 · answer #1 · answered by night_fox51 4 · 0⤊ 0⤋
convert 135 grams of Fe2O3 to moles of Fe2O3, then multiply that by 2 (because of the coefficients in the chemical equation), then convert the moles you have to the mass of the aluminum.
2007-02-05 12:00:52 · answer #2 · answered by polevaulter1000 3 · 0⤊ 0⤋
Find the molecular mass of each part of the equation using the molar masses of each element, then put them into ratios (Taking into account the molar ratios) then just solve for "x"
2007-02-05 12:05:38 · answer #3 · answered by Andrew 4 · 0⤊ 0⤋
let me get back to you i have to figure out how to show you onthe computer.
okay:
135g Fe2O3 * 1 mol Fe2O3 *2 mol Al * 27g Al
---------------------------------------------------------------------
159.6g Fe2O3 * 2 mol Fe2O3 * 1 mol Al
7290
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319.2
= aprox. 22.84 g Al
2007-02-05 12:05:06 · answer #4 · answered by Jordan Alexis 6 · 0⤊ 0⤋