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

Answer 1

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