Question

2NaClO3 (s) → 2NaCl (s) + 3O2 (g)

12.00 moles of NaClO3 will produce how many grams of O2?

MM NaClO3: 106.44 g/mol

MM NaCl: 58.44 g/mol

MM O2: 32.00 g/mol

Question 2 options:

a)

256 g of O2


b)

384 g of O2


c)

576 g of O2


d)

288 g of O2

Answer 1

Mass O2 produced : 576 g

Further explanation

Given

Reaction

2NaClO3 (s) → 2NaCl (s) + 3O2 (g)

Required

mass of O2

Solution

From the equation, mol ratio of NaClO3 : O2 = 2 : 3, so mol O2 :

= 3/2 x mol NaClO3

= 3/2 x 12 moles

= 18 moles

Mass O2 :

= mol x MW

= 18 x 32 g/mol

= 576 g