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

13.44 L

Explanation:

Given dа ta:

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