Question

Iron(III) oxide reacts with carbon monoxide according to the equation:
Fe2O3(s)+3CO(g)→2Fe(s)+3CO2(g)
A reaction mixture initially contains 22.90 g Fe2O3 and 15.78 g CO.

Once the reaction has occurred as completely as possible, what mass (in g) of the excess reactant is left?

Answer 1

Molar mass of Fe2O3,

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

= 2*55.85 + 3*16.0

= 159.7 g/mol

mass(Fe2O3)= 22.9 g

use:

number of mol of Fe2O3,

n = mass of Fe2O3/molar mass of Fe2O3

=(22.9 g)/(1.597*10^2 g/mol)

= 0.1434 mol

Molar mass of CO,

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

= 1*12.01 + 1*16.0

= 28.01 g/mol

mass(CO)= 15.78 g

use:

number of mol of CO,

n = mass of CO/molar mass of CO

=(15.78 g)/(28.01 g/mol)

= 0.5634 mol

Balanced chemical equation is:

Fe2O3 + 3 CO ---> 2 Fe + 3 CO2

1 mol of Fe2O3 reacts with 3 mol of CO

for 0.1434 mol of Fe2O3, 0.4302 mol of CO is required

But we have 0.5634 mol of CO

so, Fe2O3 is limiting reagent

we will use Fe2O3 in further calculation

According to balanced equation

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

= (3/1)*0.1434

= 0.4302 mol

mol of CO remaining = mol initially present - mol reacted

mol of CO remaining = 0.5634 - 0.4302

mol of CO remaining = 0.1332 mol

Molar mass of CO,

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

= 1*12.01 + 1*16.0

= 28.01 g/mol

use:

mass of CO,

m = number of mol * molar mass

= 0.1332 mol * 28.01 g/mol

= 3.731 g

Answer: 3.73 g