Question

The balanced equation for the reduction of iron ore to the metal using CO  is
Fe2O3 (s) + 3CO (g) ----> 2Fe (s) + 3CO2 (g)

What is the maximum mass of iron, in grams, that can be obtained from 937 g of iron(III) oxide? Mass = g Fe What mass of CO is required to react with 937 g of Fe2O3? Mass = gCO

Answer 1

1)

Molar mass of Fe2O3,

MM = 2*MM(Fe) + 3*MM(O)

= 2*55.85 + 3*16.0

= 159.7 g/mol

mass of Fe2O3 = 9.37*10^2 g

mol of Fe2O3 = (mass)/(molar mass)

= 9.37*10^2/1.597*10^2

= 5.867 mol

Balanced chemical equation is:

Fe2O3 (s) + 3CO (g) ----> 2Fe (s) + 3CO2 (g)

According to balanced equation

mol of Fe formed = (2/1)* moles of Fe2O3

= (2/1)*5.867

= 11.73 mol

Molar mass of Fe = 55.85 g/mol

mass of Fe = number of mol * molar mass

= 11.73*55.85

= 655 g

Answer: 655 g

2)

According to balanced equation

mol of CO required = (3/1)* moles of Fe2O3

= (3/1)*5.867

= 17.6 mol

Molar mass of CO,

MM = 1*MM(C) + 1*MM(O)

= 1*12.01 + 1*16.0

= 28.01 g/mol

mass of CO = number of mol * molar mass

= 17.6*28.01

= 493 g

Answer: 493 g