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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
​

Respuesta :

Neetoo

Answer:

13.44 L

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

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