If 7.50 g of NaClO3 decomposes in the presence of heat to produce 4.12 g

of NaCl, what mass of O2 gas will be produced?

2 NaClO3 (s) + 2 NaCl (s) + 3 O2 (g)

2

UO

3.38 grams of O2 will be produced

Explanation:

Step 1: Data given

Mass of NaClO3 = 7.50 grams

Mass of NaCl = 4.12 grams

Molar mass of NaClO3 = 106.44 g/mol

Molar mass of NaCl = 58.44 g/mol

Step 2: The balanced equation

2NaClO3 (s) → 2NaCl (s) + 3O2 (g)

Step 3: Calculate moles

Moles NaClO3 = mass NaClO3 / molar mass NaClO3

Moles NaClO3 = 7.50 grams / 106.44 g/mol

Moles NaClO3 = 0.0705 moles

Moles NaCl = 4.12 grams / 58.44 g/mol

Moles NaCl = 0.0705 moles

For 2 moles NaClO3 we'll have 2 moles NaCl and 3 moles O2

For 0.0705 moles NaClO3 we'll have 0.0705 moles NaCl and 3/2 * 0.0705 = 0.10575 moles O2

Step 4: Calculate mass O2

Mass O2 = moles O2 * molar mass O2

Mass O2 = 0.10575 moles * 32.0 g/mol

Mass O2 = 3.38 grams O2

3.38 grams of O2 will be produced