Question

For the reaction shown, calculate how many grams of oxygen form when each of the following completely reacts.

2KClO3(s)→2KCl(s)+3O2(g)   

1. 2.95 g KClO3

2. 0.302 g KClO3

3. 86.3 kg KClO3

4. 26.2 mg KClO3

Answer 1

1)

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 = 2.95 g

molar mass of KClO3 = 122.55 g/mol

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

= 2.95/122.55

= 0.0241 mol

Balanced chemical equation is:

2KClO3 ---> 2KCl + 3O2

According to balanced equation

mol of KCl formed = moles of KClO3

= 0.0241 mol

mass of KCl = number of mol * molar mass of KCl

= 0.0241*74.55

= 1.79 g

Answer: 1.79 g

2)

mass of KClO3 = 0.302 g

molar mass of KClO3 = 122.55 g/mol

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

= 0.302/122.55

= 2.46*10^-3 mol

Balanced chemical equation is:

2KClO3 ---> 2KCl + 3O2

According to balanced equation

mol of KCl formed = moles of KClO3

= 2.46*10^-3 mol

mass of KCl = number of mol * molar mass

= 0.0025*74.55

= 0.184 g

Answer: 0.184 g

3)

mass of KClO3 = 86.3 kg = 86300 g

molar mass of KClO3 = 122.55 g/mol

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

= 86300/122.55

= 704.2024 mol

Balanced chemical equation is:

2KClO3 ---> 2KCl + 3O2

According to balanced equation

mol of KCl formed = moles of KClO3

= 704.2024 mol

mass of KCl = number of mol * molar mass

= 704.2024*74.55

= 52498 g

= 52.5 Kg

Answer: 52.5 Kg

4)

mass of KClO3 = 0.0262 g

molar mass of KClO3 = 122.55 g/mol

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

= 0.0262/122.55

= 2.14*10^-4 mol

Balanced chemical equation is:

2KClO3 ---> 2KCl + 3O2

According to balanced equation

mol of KCl formed = moles of KClO3

= 2.14*10^-4 mol

mass of KCl = number of mol * molar mass

= 2.14*10^-4 *74.55

= 0.0159 g

Answer: 0.0159 g