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If 129.63grams of aluminum metal is reacted with767.44grams of iron (III) oxide, what mass ofaluminum oxide is likely to form?Reaction: 2Al(s) + Fe2O3(aq)--> Al2O3(aq) + 2Fe(s)

Sagot :

Step 1

The reaction must be written, completed, and balanced:

2Al(s) + Fe2O3(aq) => Al2O3(aq) + 2Fe(s)

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Step 2

Information provided:

The mass of Al = 129.63 g

The mass of Fe2O3 = 767.44 g

(because of this, the limiting reactant must be determined)

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Information needed:

The molar masses:

Al) 26.981 g/mol

Fe2O3) 159.69 g/mol

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Step 3

The limiting reactant:

2Al(s) + Fe2O3(aq) => Al2O3(aq) + 2Fe(s)

By stoichiometry => 1 mole Al = 26.981 g and 1 mole Fe2O3 = 159.69 g

Procedure:

2 x 26.981 g Al --------- 159.69 g Fe2O3

129.63 g Al ---------- X

X = 129.63 g Al x 159.69 g Fe2O3/2 x 26.981 g Al = 383.61 g

(For 129.63 g of Al, 383.61 g of Fe2O3 is needed, but there is 767.44 g of Fe2O3. Therefore, the limiting reactant is the Al and the excess is the Fe2O3)

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Step 4

The mass of Al2O3: (molar mass of Al2O3 = 101.96 g/mol)

2Al(s) + Fe2O3(aq) => Al2O3(aq) + 2Fe(s)

2 x 26.981 g Al ------- 101.96 g Al2O3

129.63 g Al ------- X = 244.93 g

Answer: mass of aluminum oxide = 244.93 g

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