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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