Consider the following redox reaction: Al(s) + Fe_2O_3 rightarrow Al_2O_3(s) + F

Question

Consider the following redox reaction: Al(s) + Fe_2O_3 rightarrow Al_2O_3(s) + Fe(l) a. Determine the mass of Fe(l) that is produced when 500.0 g of aluminum reacts with an excess of iron (III) oxide b. Determine the mass of Fe(l) that is produced when 500.0 g of iron (III) oxide reacts with an excess of aluminum c. If 500.0 g Fe_2O_3 is allowed to react with 500.0 g Al, what is the limiting reactant? d. What is the theoretical yield of fe(l) in grams? e. If 150.0 g of Fe(l) is produced in the reaction, what is the % yield> Predict whether each of the following compounds would be soluble or insoluble in water a. Ba(NO_3)_2 b. Ca_3(PO_4)_2

Explanation / Answer

Balancing the given equation:

2 Al + Fe2O3 ------> 2 Fe + Al2O3

2 mole 1 mole 2mole 1 mole

Data:

Molar mass of Al = 27 g/mol

Molar mass of Fe = 55.85 g/mol

Molar mass of Fe2O3 = 159.69 g/mol

Molar mass of Al2O3 = 102 g/mol

Solution:

(a) given 500 g of Al

No. of moles of Al = Mass of Al / Molar mass of Al = 500 g / 27 g/mol = 18.52 moles

given ferrous oxide is in excess hence Al is limiting reactant

According to reaction stoichometry 2 mole Al will give 2 mole of Fe

18.52 mole of Al will give 18.52 mole of Fe

Mass of Fe formed = No. of moles of Fe * Molar mass of Fe = 18.52 moles * 55.85 g/mol = 1034.26 g

Answer (a) 1034.26 g of Fe

(b) given 500 g of Fe2O3

No. of moles of Fe2O3 = Mass of Fe2O3 / Molar mass of Fe2O3 = 500 g / 159.69 g/mol = 3.13 moles

given Al is in excess hence Fe2O3 is limiting reactant

According to reaction stoichometry 1 mole Fe2O3 will give 2 mole of Fe

3.13 mole of Fe2O3 will give 6.26 mole of Fe

Mass of Fe formed = No. of moles of Fe * Molar mass of Fe = 6.26 moles * 55.85 g/mol = 349.74 g

Answer (b) 349.74 g of Fe

(c) if 500 g of Fe2O3 and 500 g of Al reacts

then there will 3.13 moles of Fe2O3 and 18.52 moles of Al

according to stoichometry

for 3.13 moles of Fe2O3 there should be 6.26 moles of Al

But Al is in excess hence Fe2O3 is the limiting reactant

Answer (c)  Fe2O3 is the limiting reactant

so 6.26 moles of Al react with 3.13 moles of Fe2O3 to give 6.26 moles of Fe

Answer (d) so theoretical yield is 6.26 moles of Fe or 349.62 g of Fe

Mass of Theoretical yield = No. of moles * molar mass of Fe = 6.26 * 55.85 = 349,62 g

if 150 g of Fe is produced

No. of moles of Fe produced = Mass of Fe / molar mass of Fe = 150 g / 55.85 g/mol = 2.685 moles

so % yield = (mass of Actual yield / Mass of Theoretical Yield) * 100 % = 150 g / 349.62 g * 100%

% yield = 42.9 % Answer

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