Question

Pure oxygen gas was first prepared by heating mercury (ii) oxide, hgo: 2hgo(s)→2hg(l)+o2(g) what volume (in l) of oxygen at stp is released by heating 10.36 g of hgo?

Answer 1

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 2

Answer:

The volume occupied by the oxygen gas is 0.5353 L.

Explanation:

$HgO(s)\rightarrow 2Hg(l)+O_2(g)$

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

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

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

$\frac{1}{2}\times 0.0478=0.0239 mol$

At STP, the 1 mol gas occupies = 22.4 L

Then 0.0239 mol of oxygen at STP will occupy volume =$0.0239\times 22.4 L=0.5353 L$

The volume occupied by the oxygen gas is 0.5353 L.