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Answer: Moles HgO = 10.36 g/ 216.59 g/mol= 0.048 the ratio between HgO and O2 is 2 : 1 moles O2 formed = 0.048/ 2= 0.024 at STP one mole of any gas occupies 22.4 L Volume O2 = 0.024 x 22.4 L/mol = 0.54 L

Answer:

The volume occupied by the oxygen gas is 0.5353 L.

Explanation:

[tex]HgO(s)\rightarrow 2Hg(l)+O_2(g)[/tex]

Moles mercury(II) oxide =[tex]\frac{10.36 g}{216.59 g/mol}0.0478 mol[/tex]

According to reaction, 2 mol of  mercury(II) oxide gives 1 mol of oxygen.

Then 0.0478 mol of mercury(II) oxide will give:

[tex]\frac{1}{2}\times 0.0478=0.0239 mol[/tex]

At STP, the 1 mol gas occupies = 22.4 L

Then 0.0239 mol of oxygen at STP will occupy volume =[tex]0.0239\times 22.4 L=0.5353 L[/tex]

The volume occupied by the oxygen gas is 0.5353 L.

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