Questions: Solving for a gaseous reactant 3 / 5 Kassi 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 23.0 mL of dioxygen gas are produced by this reaction, at a temperature of 130.0°C and pressure of exactly 1 atm. Calculate the mass of mercury(II) oxide that must have reacted. Round your answer to 3 significant digits. g

Solution Steps

The decomposition of mercury(II) oxide (\(\text{HgO}\)) into liquid mercury (\(\text{Hg}\)) and gaseous dioxygen (\(\text{O}_2\)) can be represented by the following balanced chemical equation:

\[ 2 \, \text{HgO (s)} \rightarrow 2 \, \text{Hg (l)} + \text{O}_2 \, \text{(g)} \]

To find the mass of mercury(II) oxide that reacted, we first need to determine the number of moles of dioxygen gas produced. We can use the ideal gas law:

\[ PV = nRT \]

Where:

• \(P = 1 \, \text{atm}\)
• \(V = 23.0 \, \text{mL} = 0.0230 \, \text{L}\)
• \(R = 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1}\)
• \(T = 130.0^{\circ} \text{C} = 403.15 \, \text{K}\)

Rearranging the ideal gas law to solve for \(n\) (moles of \(\text{O}_2\)):

\[ n = \frac{PV}{RT} = \frac{(1 \, \text{atm})(0.0230 \, \text{L})}{(0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1})(403.15 \, \text{K})} \]

Calculating \(n\):

\[ n \approx 0.000698 \, \text{mol} \]

From the balanced equation, 2 moles of \(\text{HgO}\) produce 1 mole of \(\text{O}_2\). Therefore, the moles of \(\text{HgO}\) that reacted are:

\[ n_{\text{HgO}} = 2 \times n_{\text{O}_2} = 2 \times 0.000698 \, \text{mol} = 0.001396 \, \text{mol} \]

The molar mass of \(\text{HgO}\) is approximately \(216.59 \, \text{g/mol}\). Thus, the mass of \(\text{HgO}\) is:

\[ \text{mass of HgO} = n_{\text{HgO}} \times \text{molar mass of HgO} = 0.001396 \, \text{mol} \times 216.59 \, \text{g/mol} \]

Calculating the mass:

\[ \text{mass of HgO} \approx 0.302 \, \text{g} \]

Final Answer