Question

2KMnO4+16HCl -> 5Cl2+2KCl+2MnCl2+8H20

A. How many moles of water will be produced when 5.0 moles of KMnO4 are consumed?

B. What is the maximum mass of Cl2 that can be produced by reacting 88.0g KMnO4?

Answer 1

A)

2KMnO4+16HCl -> 5Cl2+2KCl+2MnCl2+8H20

from balanced reaction,

moles of H2O = (8/2)*moles of KMnO4

moles of H2O = 4*moles of KMnO4

moles of H2O = 4*5.0 moles

moles of H2O = 20 moles

Answer: 20 moles

B)

mass of KMnO4 = 88.0 g

molar mass of KMnO4 = 158 g/mol

mol of KMnO4 = (mass)/(molar mass)

= 88/158

= 0.557 mol

According to balanced equation

mol of Cl2 formed = (5/2)* moles of KMnO4

= (5/2)*0.557

= 1.392 mol

mass of Cl2 = number of mol * molar mass

= 1.392*70.9

= 98.7 g

Answer: 98.7 g

answered by: anonymous