Question

A 20.0 g sample of mercury(II) oxide (HgO, M = 216.6) is heated strongly, causing it to decompose to metallic Hg and O2 gas. What volume of O2 gas is produced (measured at STP)?

Answer 1

Answer:The volume of oxygen gas at STP will be 1.1278 L.

Explanation:

$2HgO\rightarrow 2Hg+O_2$

Moles of HgO = $\fac{\text{mass of HgO}}{\text{molar mass of HgO}}=\frac{20.0 g}{216.6 g/mol}=0.0923 mol$

According to reaction ,2 moles of HgO gives one mole of oxygen gas then , 0.0923 mol of HgO will give = $\frac{1}{2}\times 0.0923 moles$ of oxygen gas that is 0.0461 mol .

At STP, the pressure is at 1 atm , temperature is 298 K, so the volume occupied by 0.0461 moles of oxygen at STP will be :

$V=\frac{nRT}{P}=\frac{0.0461\times 0.0821atm L/mol K\\times 298 K}{1 atm}=1.1278 L$

The volume of oxygen gas at STP will be 1.1278 L.

Answer 2

Answer:

A) 1.03 L

Explanation:

20.0g/216.6 (g/mol HgO) = 0.0923 mol HgO

0.0923 mol HgO / (1 mol O2 x 2 mol HgO) = 0.04615 mol O2

0.04615 mol O2 x 22.4 L/1 mol = 1.03 L O2 produced