Question

For the reaction

2KClO3(s)→2KCl(s)+3O2(g)

calculate how many grams of oxygen form when each quantity of reactant completely reacts.

a. 2.91 gKClO3

b. 0.400 gKClO3

c. 87.5 kgKClO3

d. 25.9 mgKClO3

Answer 1

a)

Molar mass of KClO3,

MM = 1*MM(K) + 1*MM(Cl) + 3*MM(O)

= 1*39.1 + 1*35.45 + 3*16.0

= 122.55 g/mol

mass of KClO3 = 2.91 g

mol of KClO3 = (mass)/(molar mass)

= 2.91/1.226*10^2

= 2.375*10^-2 mol

According to balanced equation

mol of O2 formed = (3/2)* moles of KClO3

= (3/2)*2.375*10^-2

= 3.562*10^-2 mol

Molar mass of O2 = 32 g/mol

mass of O2 = number of mol * molar mass

= 3.562*10^-2*32

= 1.14 g

Answer: 1.14 g

b)

Molar mass of KClO3,

MM = 1*MM(K) + 1*MM(Cl) + 3*MM(O)

= 1*39.1 + 1*35.45 + 3*16.0

= 122.55 g/mol

mass of KClO3 = 0.4 g

mol of KClO3 = (mass)/(molar mass)

= 0.4/1.226*10^2

= 3.264*10^-3 mol

According to balanced equation

mol of O2 formed = (3/2)* moles of KClO3

= (3/2)*3.264*10^-3

= 4.896*10^-3 mol

Molar mass of O2 = 32 g/mol

mass of O2 = number of mol * molar mass

= 4.896*10^-3*32

= 0.1567 g

Answer: 0.157 g

c)

Molar mass of KClO3,

MM = 1*MM(K) + 1*MM(Cl) + 3*MM(O)

= 1*39.1 + 1*35.45 + 3*16.0

= 122.55 g/mol

mass of KClO3 = 8.75*10^4 g

mol of KClO3 = (mass)/(molar mass)

= 8.75*10^4/1.226*10^2

= 7.14*10^2 mol

According to balanced equation

mol of O2 formed = (3/2)* moles of KClO3

= (3/2)*7.14*10^2

= 1.071*10^3 mol

Molar mass of O2 = 32 g/mol

mass of O2 = number of mol * molar mass

= 1.071*10^3*32

= 3.427*10^4 g

Answer: 3.43*10^4 g

d)

Molar mass of KClO3,

MM = 1*MM(K) + 1*MM(Cl) + 3*MM(O)

= 1*39.1 + 1*35.45 + 3*16.0

= 122.55 g/mol

mass of KClO3 = 2.59*10^-2 g

mol of KClO3 = (mass)/(molar mass)

= 2.59*10^-2/1.226*10^2

= 2.113*10^-4 mol

According to balanced equation

mol of O2 formed = (3/2)* moles of KClO3

= (3/2)*2.113*10^-4

= 3.17*10^-4 mol

Molar mass of O2 = 32 g/mol

mass of O2 = number of mol * molar mass

= 3.17*10^-4*32

= 1.014*10^-2 g

Answer: 1.01*10^-2 g