Question

The great French chemist Antoine Lavoisier discovered the Law of Conservation of Mass in part by doing a famous experiment in 1775. In this experiment Lavoisier found that mercury(II) oxide, when heated, decomposed into liquid mercury and an invisible and previously unknown substance: oxygen gas.

1. Write a balanced chemical equation, including physical state symbols, for the decomposition of solid mercury(II) oxide (HgO) into liquid mercury and gaseous dioxygen.
2. Suppose 50.0ml of dioxygen gas are produced by this reaction, at a temperature of 90°C and pressure of exactly 1atm. Calculate the mass of mercury(II) oxide that must have reacted. Be sure your answer has the correct number of significant digits.

Answer 1

Answer:

(a) $2HgO_{(s)}\rightarrow 2Hg_{(l)}+O_2_{(g)}$

(b) 0.726 g

Explanation:

(a)

The balanced chemical equation is shown below as:-

$2HgO_{(s)}\rightarrow 2Hg_{(l)}+O_2_{(g)}$

(b)

Given that:

Temperature = 90 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T₁ = (90 + 273.15) K = 363.15 K

V = 50.0 mL = 0.05 L ( 1 mL = 0.001 L )

Pressure = 1 atm

Using ideal gas equation as:

$PV=nRT$

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L atm/ K mol  

Applying the equation as:

1 atm × 0.05 L = n ×0.0821 L atm/ K mol  × 363.15 K

⇒n = 0.001677 mol

Thus, moles of $O_2$ = 0.001677 mol

According to the reaction shown above,

1 mole of $O_2$ is produced when 2 moles of mercury(II) oxide are reacted

0.001677 mole of $O_2$ is produced when $2\times 0.001677$ moles of mercury(II) oxide are reacted

Moles of mercury(II) oxide = 0.003354 moles

Molar mass of mercury(II) oxide = 216.59 g/mol

Mass of mercury(II) oxide = $Moles\times Molar\ mass$ =  $0.003354\times 216.59\ g=0.726\ g$

0.726 g is the mass of mercury(II) oxide that must have reacted.