Question

How many milliliters of oxygen gas at STP are released from heating \(2.50 \mathrm{~g}\) of mercuric oxide? $$2 \mathrm{HgO}(s) \longrightarrow 2 \mathrm{Hg}(l)+\mathrm{O}_{2}(g)$$

Short answer

129 mL of \( \mathrm{O}_{2} \) gas are released at STP.

Step-by-step solution

Determine the molar mass of HgO

Start by calculating the molar mass of mercuric oxide, HgO. The molar mass of Hg is approximately 200.59 g/mol, and the molar mass of O is about 16.00 g/mol. Therefore, the molar mass of HgO is \( 200.59 + 16.00 = 216.59 \) g/mol.

Calculate the moles of HgO present

Using the given mass of HgO (2.50 g) and its molar mass, we find the number of moles of HgO. Calculate it using the formula: \( \text{moles} = \frac{\text{mass}}{\text{molar}\ \text{mass}} \). Thus, \( \text{moles of HgO} = \frac{2.50\ \mathrm{g}}{216.59\ \mathrm{g/mol}} \approx 0.01154 \ \text{mol} \).

Use the balanced equation to find moles of O extsubscript{2} released

The balanced chemical equation shows that 2 moles of HgO decompose to release 1 mole of \( \mathrm{O}_2 \). Thus, \( \frac{1}{2} \) mole of \( \mathrm{O}_2 \) is released per mole of HgO. For 0.01154 moles of HgO, the moles of \( \mathrm{O}_2 \) released are \( 0.01154 \times \frac{1}{2} = 0.00577 \ \text{mol} \).

Calculate the volume of \( \mathrm{O}_{2} \) gas at STP

At Standard Temperature and Pressure (STP), 1 mole of any gas occupies 22.4 liters. Therefore, the volume \( V \) of \(0.00577\ \text{mol} \) of \( \mathrm{O}_{2} \) is \( 0.00577 \times 22.4 \ \mathrm{L}\). Converting this to milliliters (1 L = 1000 mL) gives \( V = 0.129248 \times 1000 = 129.248 \ \mathrm{mL} \).

Round the final volume

Since the original mass of HgO was given to 3 significant figures (2.50 g), round the calculated volume of \( \mathrm{O}_{2} \) gas to 3 significant figures. Therefore, the final result is \( 129 \ \mathrm{mL} \).

Key concepts

Understanding Stoichiometry

Stoichiometry is a fundamental concept in chemistry, involving the calculation of reactants and products in chemical reactions. It serves as a guide to understanding how quantities of substances relate to each other. When dealing with stoichiometry, we use balanced chemical equations to determine the proportions in which substances react or are produced.
In the given exercise, the balanced equation is \(2 \mathrm{HgO}(s) \rightarrow 2 \mathrm{Hg}(l)+\mathrm{O}_{2}(g)\). This equation tells us that two moles of mercuric oxide produce one mole of oxygen gas and two moles of liquid mercury.
A key aspect of stoichiometry is conversion between moles of different substances using mole ratios derived from balanced equations, allowing us to calculate amounts of products formed from given reactants.

Molar Mass: The Gateway to Moles

Molar mass acts as a bridge between the mass of a substance and the number of moles, a fundamental unit in stoichiometry. The molar mass of a compound is the sum of the atomic masses of all the atoms in its formula, expressed in grams per mole.
For example, in mercuric oxide (HgO), the atomic mass of mercury (Hg) is approximately 200.59 g/mol and that of oxygen (O) is about 16.00 g/mol. Therefore, the molar mass of HgO is calculated by adding these two values, resulting in 216.59 g/mol.
With the molar mass known, you can find the number of moles in a given sample by dividing the sample's mass by the molar mass. This calculation converts mass into moles, which is essential for further stoichiometric computations.

Gas Volume at Standard Temperature and Pressure (STP)

Understanding gas volume calculations at Standard Temperature and Pressure (STP) is crucial when dealing with gaseous products in reactions. STP is defined as a temperature of 0°C (273.15 K) and a pressure of 1 atmosphere (atm), conditions under which one mole of any ideal gas occupies 22.4 liters.
In the exercise, to find the volume of oxygen gas produced, we first calculate the moles of \( \mathrm{O}_2 \) formed from the decomposition of HgO. After finding the moles of gas, we multiply by the molar volume at STP (22.4 L/mol). This straightforward relationship allows us to convert directly from moles to volume, giving us the answer in liters, which can then be easily converted to milliliters for precision.
Remember, this conversion is possible only because we are working under STP conditions.

Basics of Chemical Reactions

Chemical reactions involve the transformation of reactants into products, showcasing the dynamic nature of matter. A chemical reaction is usually represented by a balanced chemical equation, which reflects the conservation of atoms and mass.
In the given reaction, mercuric oxide (\( \mathrm{HgO} \)) decomposes into elemental mercury (\( \mathrm{Hg} \)) and oxygen gas (\( \mathrm{O}_2 \)), demonstrating a decomposition reaction type.
Balancing chemical equations is crucial because it ensures that the number of atoms for each element is conserved throughout the reaction. This balance provides the necessary mole ratios for stoichiometric calculations, which are used to predict quantities of products formed from known amounts of reactants.
Understanding these fundamentals equips you with the knowledge to tackle stoichiometric problems systematically.