Question

Answer: 2.3. A sample of mercury(II) oxide is placed in a 5.00 L evacuated container and heated until it decomposes entirely to mercury metal and oxygen gas. The container is then cooled to 25°C. One now finds that the gas pressure inside the container is 1.73 atm. What mass of mercury(II)oxide was originally placed into the container? Answer:

Answer 1

Given

Volume of oxygen gas = volume of container = 5.00 L

Temperature of gas = 25 + 273 = 298 K

Pressure of oxygen gas = 1.73 atm

Mass of HgO originally placed into the container = ?

Decomposition reaction of HgO is, 2 HgO (s)  $\overset{\Delta }{\rightarrow}$ 2 Hg (s) + O ₂ (g)

From reaction, 2 mol HgO $\equiv$ 2 mol Hg $\equiv$ 1 mol O ₂

First calculate no. of moles of HgO from the no. of moles of oxygen produced in the reaction.

We have, P V = n R T

where, P is a pressure of a gas, V is a volume of a gas, n is no. of moles of gas, R is a gas constant and T is temperature of gas.

$\therefore$ No. of moles of oxygen gas ( n ) = P V / R T

$\therefore$ No. of moles of oxygen gas ( n ) = ( 1.73 atm $\times$ 5.00 L ) / (  0.082057 L atm mol ^-1 K ^-1 298 K )

$\therefore$ No. of moles of oxygen gas ( n ) = 0.3537

From reaction, 1 mol O ₂$\equiv$2 mol HgO

$\therefore$ 0.3537 mol O ₂$\equiv$ ( 2 $\times$ 0.3537 / 1 ) mol HgO

$\therefore$ 0.3537 mol O ₂$\equiv$ 0.7074 mol HgO

We have relation, no. of moles = Mass / Molar mass

$\therefore$ Mass = No. of moles $\times$ Molar mass

$\therefore$ Mass of HgO = No. of moles of HgO $\times$ Molar mass of HgO

Molar mass of HgO = 200.59 + 16.00 = 216.59 g/ mol

$\therefore$ Mass of HgO = 0.7074 mol  $\times$ 216.59 g/ mol = 153 g

ANSWER : Mass of HgO originally placed into the container = 153 g