If 5.0g of KCLO3 is decomposed, what volume of O2 is produced at STP?

V = 1.4 L

Explanation:

Given data:

Mass of KClO₃  = 5 g

Volume of oxygen produced = ?

Solution:

Chemical equation:

2KClO₃ → 2KCl +3O₂

Number of moles of KClO₃:

Number of moles = mass/ molar mass

Number of moles = 5 g / 122.55 g/mol

Number of moles = 0.041

Now we will compare the moles of KClO₃ with oxygen,

                    KClO₃         :          O₂

                       2              :           3

                      0.041       :      3/2× 0.041 = 0.062

Volume of oxygen at STP:

Standard pressure = 1 atm

Standard temperature = 273 K

PV = nRT

V = nRT/P

V = 0.062 × 0.0821 atm.L/mol.K × 273 K / 1 atm

V = 1.4 atm.L / 1atm

V = 1.4 L

the answer is d. 74.02