Question

Pure oxygen was first made by heating mercury(II) oxide: \(\mathrm{Hg} \mathrm{O} \stackrel{\text { Heat }}{\longrightarrow} \mathrm{Hg}+\mathrm{O}_{2} \quad\) Unbalanced (a) Balance the equation. (b) How many grams of mercury and how many grams of oxygen are formed from \(45.5 \mathrm{~g}\) of \(\mathrm{HgO}\) ? (c) How many grams of \(\mathrm{HgO}\) would you need to obtain \(33.3 \mathrm{~g}\) of \(\mathrm{O}_{2} ?\)

Short answer

(a) Balanced equation: \(2\mathrm{HgO} \rightarrow 2\mathrm{Hg} + \mathrm{O}_2\). (b) 42.12 g of Hg and 3.36 g of O2 are formed. (c) 450.8 g of HgO is needed.

Step-by-step solution

Balance the Chemical Equation

The unbalanced chemical equation is: \( \text{HgO} \rightarrow \text{Hg} + \text{O}_2 \). To balance the equation, start by counting the number of atoms of each element on both sides. The oxygen and mercury atoms must be balanced. By placing a coefficient of 2 in front of \( \text{HgO} \) and \( \text{Hg} \), the equation becomes: \[ 2\text{HgO} \rightarrow 2\text{Hg} + \text{O}_2 \] which balances the equation with 2 oxygen atoms and 2 mercury atoms on each side.

Calculate Molar Masses

Calculate the molar mass of each compound involved using the periodic table: \( \text{HgO} \) has a molar mass of approximately 216.6 g/mol (Hg: 200.59, O: 16.00). \( \text{Hg} \) has a molar mass of 200.59 g/mol. \( \text{O}_2 \) has a molar mass of 32.00 g/mol.

Mass of Products from 45.5 g of HgO

Based on the balanced equation, 2 mol of \( \text{HgO} \) produce 2 mol of \( \text{Hg} \) and 1 mol of \( \text{O}_2 \). First, calculate the number of moles of \( \text{HgO} \) in 45.5 g: \[ \text{moles of HgO} = \frac{45.5 \text{ g}}{216.6 \text{ g/mol}} \approx 0.210 \text{ mol} \]Using stoichiometry: - \( 2 \text{ mol HgO} \rightarrow 2 \text{ mol Hg} \)- \( 0.210 \text{ mol HgO} \rightarrow 0.210 \text{ mol Hg} \)Mass of \( \text{Hg} \) formed: \[ = 0.210 \text{ mol} \times 200.59 \text{ g/mol} \approx 42.12 \text{ g} \]- \( 2 \text{ mol HgO} \rightarrow 1 \text{ mol O}_2 \)- \( 0.210 \text{ mol HgO} \rightarrow 0.105 \text{ mol O}_2 \)Mass of \( \text{O}_2 \) formed:\[ = 0.105 \text{ mol} \times 32.00 \text{ g/mol} \approx 3.36 \text{ g} \]

Mass of HgO Required for 33.3 g of O2

From the balanced equation, \( 2 \text{ mol HgO} \rightarrow 1 \text{ mol O}_2 \).Calculate moles of \( \text{O}_2 \) in 33.3 g: \[ \text{moles of } O_2 = \frac{33.3 \text{ g}}{32.00 \text{ g/mol}} \approx 1.041 \text{ mol} \]Using stoichiometry:- \( 1 \text{ mol O}_2 \) produced by \( 2 \text{ mol HgO} \)- \( 1.041 \text{ mol O}_2 \) produced by \( 2.082 \text{ mol HgO} \)Mass of \( \text{HgO} \) required:\[ = 2.082 \text{ mol} \times 216.6 \text{ g/mol} \approx 450.8 \text{ g} \]

Key concepts

Stoichiometry

Stoichiometry is a crucial concept in chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It allows chemists to predict how much of each substance is needed or produced in a reaction. Let's illustrate it with mercury(II) oxide decomposition.
To balance the chemical equation, we use stoichiometry to ensure the number of atoms for each element is the same on both sides of the equation. In this case, for every 2 moles of HgO decomposed, we produce 2 moles of Hg and 1 mole of O extsubscript{2}.
This relationship helps calculate how much product we can expect from a certain amount of reactant. This is a stoichiometric calculation that involves finding mole ratios and applying them to find the masses involved in the reaction.

Molar Mass Calculations

Molar mass calculations are essential for converting between grams and moles, which are fundamental units in chemistry. Knowing the molar mass of substances allows us to apply stoichiometry effectively.
For example, the molar mass of HgO is approximately 216.6 g/mol, calculated by adding the atomic masses of mercury (200.59 g/mol) and oxygen (16.00 g/mol).

• Calculating the molar mass of Hg: 200.59 g/mol.
• For O extsubscript{2}, the molar mass is 32.00 g/mol since oxygen's atomic mass is 16.00 g/mol, and the molecule consists of two oxygen atoms.

By using these molar masses, we can convert grams to moles, or vice versa, to understand how much of each substance is involved in the reaction.

Chemical Reactions

Chemical reactions are processes in which substances, known as reactants, are transformed into other substances, known as products.
In our example, the chemical reaction is the decomposition of mercury(II) oxide (HgO) into mercury (Hg) and oxygen (O extsubscript{2}). The unbalanced equation is given, and balancing it is the first step.
This decomposition reaction involves breaking the HgO compound into simpler substances. By balancing the equation, we ensure the conservation of mass, meaning that the same number of each type of atom is present in both reactants and products.

• An unbalanced equation: HgO → Hg + O extsubscript{2}
• Balanced equation: 2HgO → 2Hg + O extsubscript{2}

This shows how stoichiometry aids in maintaining balance in chemical reactions.

Mercury(II) Oxide Decomposition

Mercury(II) oxide decomposition is a specific type of chemical reaction that involves the breakdown of a compound into simpler substances, in this case, breaking down HgO into Hg and O extsubscript{2}.
When heat is applied to HgO, it decomposes according to the balanced equation: 2HgO → 2Hg + O extsubscript{2}. This reaction was historically significant as it was one of the first ways to obtain pure oxygen.
Through this reaction, the stoichiometric relationships become evident. For instance, for every 2 moles of HgO decomposed, 1 mole of O extsubscript{2} is generated along with 2 moles of Hg.
This illustrates the chemical change in the decomposition, diverging from the initial reactant into distinct elements.