Question

2 KMnO4 + 16 HCl = 5 Cl2 + 2 KCl + 2 MnCl2 + 8 H2O

If you react 22.7 grams KMnO4 with 48.6 grams HCl

a) how much KCl will be produced?

b) How much of the excess reactant will remain after the reaction has come to an end?

Answer 1

1)

Molar mass of KMnO4,

MM = 1*MM(K) + 1*MM(Mn) + 4*MM(O)

= 1*39.1 + 1*54.94 + 4*16.0

= 158.04 g/mol

mass(KMnO4)= 22.7 g

use:

number of mol of KMnO4,

n = mass of KMnO4/molar mass of KMnO4

=(22.7 g)/(1.58*10^2 g/mol)

= 0.1436 mol

Molar mass of HCl,

MM = 1*MM(H) + 1*MM(Cl)

= 1*1.008 + 1*35.45

= 36.458 g/mol

mass(HCl)= 48.6 g

use:

number of mol of HCl,

n = mass of HCl/molar mass of HCl

=(48.6 g)/(36.46 g/mol)

= 1.333 mol

Balanced chemical equation is:

2 KMnO4 + 16 HCl = 5 Cl2 + 2 KCl + 2 MnCl2 + 8 H2O

2 mol of KMnO4 reacts with 16 mol of HCl

for 0.1436 mol of KMnO4, 1.149 mol of HCl is required

But we have 1.333 mol of HCl

so, KMnO4 is limiting reagent

we will use KMnO4 in further calculation

Molar mass of KCl,

MM = 1*MM(K) + 1*MM(Cl)

= 1*39.1 + 1*35.45

= 74.55 g/mol

According to balanced equation

mol of KCl formed = (2/2)* moles of KMnO4

= (2/2)*0.1436

= 0.1436 mol

use:

mass of KCl = number of mol * molar mass

= 0.1436*74.55

= 10.71 g

Answer: 10.7 g

2)

According to balanced equation

mol of HCl reacted = (16/2)* moles of KMnO4

= (16/2)*0.1436

= 1.149 mol

mol of HCl remaining = mol initially present - mol reacted

mol of HCl remaining = 1.333 - 1.149

mol of HCl remaining = 0.184 mol

Molar mass of HCl,

MM = 1*MM(H) + 1*MM(Cl)

= 1*1.008 + 1*35.45

= 36.458 g/mol

use:

mass of HCl,

m = number of mol * molar mass

= 0.184 mol * 36.46 g/mol

= 6.707 g

Answer: 6.71 g