Question

Iron reacts with chlorine gas according to the following balanced chemical equation:

2Fe(s)+3Cl2(g)→2FeCl3(s)

A) What mass of iron(III) chloride is produced when 101.08 g iron reacts with 185.42 g chlorine gas?

B) What is the percent yield of iron(III) chloride if 139.13 g of iron(III) chloride is actually obtained?

Answer 1

A)

Molar mass of Fe = 55.85 g/mol

mass(Fe)= 101.08 g

use:

number of mol of Fe,

n = mass of Fe/molar mass of Fe

=(1.011*10^2 g)/(55.85 g/mol)

= 1.81 mol

Molar mass of Cl2 = 70.9 g/mol

mass(Cl2)= 185.42 g

use:

number of mol of Cl2,

n = mass of Cl2/molar mass of Cl2

=(1.854*10^2 g)/(70.9 g/mol)

= 2.615 mol

Balanced chemical equation is:

2 Fe + 3 Cl2 ---> 2 FeCl3 +

2 mol of Fe reacts with 3 mol of Cl2

for 1.81 mol of Fe, 2.715 mol of Cl2 is required

But we have 2.615 mol of Cl2

so, Cl2 is limiting reagent

we will use Cl2 in further calculation

Molar mass of FeCl3,

MM = 1*MM(Fe) + 3*MM(Cl)

= 1*55.85 + 3*35.45

= 162.2 g/mol

According to balanced equation

mol of FeCl3 formed = (2/3)* moles of Cl2

= (2/3)*2.615

= 1.743 mol

use:

mass of FeCl3 = number of mol * molar mass

= 1.743*1.622*10^2

= 283 g

Answer: 283 g

b)

% yield = actual mass*100/theoretical mass

= 1.391*10^2*100/2.828*10^2

= 49.2%

Answer: 49.2 %