Question

8. How many liters of oxygen gas will be produced by the decomposition of

100.0 grams of sodium nitrate? (Note: 1 mole of any gas at STP will have a
vol= 22.4 liters.)
2 NaNO3 → 2NaNO2 + O2
​

Answer 1

13.44 L

Step-by-step explanation:

Given data:

Volume of oxygen produced = ?

Mass of sodium nitrate = 100 g

Solution:

Chemical equation:

2NaNO₃ → 2NaNO₂ + O₂

Number of moles of sodium nitrate:

Number of moles = mass/ molar mass

Number of moles = 100 g/ 85 g/mol

Number of moles = 1.2 mol

Now we will compare the moles of NaNO₃ with oxygen

NaNO₃ : O₂

2 : 1

1.2 : 1/2 × 1.2 = 0.6 mol

Volume:

one mole = 22.4 L

0.6 mol × 22.4 = 13.44 L