In: Chemistry
Iron(III) oxide reacts with carbon monoxide according to the
equation:
Fe2O3(s)+3CO(g)→2Fe(s)+3CO2(g)
A reaction mixture initially contains 22.25 g Fe2O3and 14.18 g
CO.
Once the reaction has occurred as completely as possible, what mass (in g) of the excess reactant is left?
Molar mass of Fe2O3,
MM = 2*MM(Fe) + 3*MM(O)
= 2*55.85 + 3*16.0
= 159.7 g/mol
mass(Fe2O3)= 22.25 g
use:
number of mol of Fe2O3,
n = mass of Fe2O3/molar mass of Fe2O3
=(22.25 g)/(1.597*10^2 g/mol)
= 0.1393 mol
Molar mass of CO,
MM = 1*MM(C) + 1*MM(O)
= 1*12.01 + 1*16.0
= 28.01 g/mol
mass(CO)= 14.18 g
use:
number of mol of CO,
n = mass of CO/molar mass of CO
=(14.18 g)/(28.01 g/mol)
= 0.5062 mol
Balanced chemical equation is:
Fe2O3 + 3 CO ---> 2 Fe + 3 CO2
1 mol of Fe2O3 reacts with 3 mol of CO
for 0.1393 mol of Fe2O3, 0.418 mol of CO is required
But we have 0.5062 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)*0.1393
= 0.418 mol
mol of CO remaining = mol initially present - mol reacted
mol of CO remaining = 0.5062 - 0.418
mol of CO remaining = 8.828*10^-2 mol
use:
mass of CO,
m = number of mol * molar mass
= 8.828*10^-2 mol * 28.01 g/mol
= 2.473 g
Answer: 2.47 g