Question

The weight of \(\mathrm{MnO}_{2}\) required to produce \(1.78\) litres of chlorine gas at STP according to the reaction \(\mathrm{MnO}_{2}+4 \mathrm{HCl} \longrightarrow \mathrm{MnCl}_{2}+2 \mathrm{H}_{2} \mathrm{O}+\mathrm{Cl}_{2} \mathrm{is}\) (1) \(6.905 \mathrm{~g}\) (2) \(5.905 \mathrm{~g}\) (3) \(6.509 \mathrm{~g}\) (4) \(6.059 \mathrm{~g}\)

Short answer

6.905 g

Step-by-step solution

- Determine Moles of Chlorine Gas

Use the ideal gas law to determine the number of moles of chlorine gas produced. At STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 liters. Given volume of \(\text{Cl}_2\) is 1.78 liters, we can find the moles by dividing this volume by 22.4 liters per mole. \[\text{Moles of } \text{Cl}_2 = \frac{1.78 \text{ L}}{22.4 \text{ L/mol}} = 0.0795 \text{ mol}\text{ Cl}_2\text{.}\]

- Calculate Moles of \( \text{MnO}_2 \) Required

From the balanced chemical equation \( \text{MnO}_2 + 4 \text{HCl} \rightarrow \text{MnCl}_2 + 2 \text{H}_2\text{O} + \text{Cl}_2 \), it is evident that 1 mole of \( \text{MnO}_2 \) produces 1 mole of \( \text{Cl}_2 \). Therefore, the moles of \( \text{MnO}_2 \) required is equal to the moles of \( \text{Cl}_2 \) produced. We have already calculated the moles of \( \text{Cl}_2 \) to be 0.0795 moles. So, moles of \( \text{MnO}_2 \) required is also 0.0795 moles.

- Find Molar Mass of \( \text{MnO}_2 \)

Calculate the molar mass of \( \text{MnO}_2 \). The atomic masses of Mn and O are approximately 54.94 g/mol and 16.00 g/mol, respectively. Therefore, the molar mass of \( \text{MnO}_2 \) is \[ \text{Molar mass of } \text{MnO}_2 = 54.94 + 2 \times 16.00 = 86.94 \text{ g/mol} \]

- Calculate Mass of \( \text{MnO}_2 \) Required

The mass of a substance can be found by multiplying the number of moles by the molar mass. Using our results from steps 1 and 3, we find \[ \text{Mass of } \text{MnO}_2 = 0.0795 \text{ mol} \times 86.94 \text{ g/mol} = 6.905 \text{ g} \]

Key concepts

Chemical Reactions

In chemistry, a chemical reaction is a process where reactants are transformed into products. Reactants are the starting materials, and products are the substances formed as a result of the reaction. Here, the given reaction is: \ \( \text{MnO}_2 + 4 \text{HCl} \rightarrow \text{MnCl}_2 + 2 \text{H}_2\text{O} + \text{Cl}_2 \ \).
This balanced chemical equation shows that one mole of manganese dioxide (MnO\text{_}2) reacts with four moles of hydrochloric acid (HCl) to produce one mole of manganese chloride (MnCl\text{_}2), two moles of water (H\text{_}2O), and one mole of chlorine gas (Cl\text{_}2).
Understanding the mole ratio is essential because it allows you to relate the quantities of different substances reacting or being produced.

Molar Mass Calculations

Molar mass is the mass of one mole of a substance, measured in grams per mole (g/mol). It is determined by adding up the atomic masses of all atoms in a molecular formula. For example, the molar mass of manganese dioxide (\text{MnO}\text{_}2) is calculated as follows:
\ \( \text{Molar mass of MnO}_2 = 54.94 \text{ g/mol} \text{ (Mn)} + 2 \times 16.00 \text{ g/mol} \text{ (O)} = 86.94 \text{ g/mol} \ \).
The atomic mass of manganese (Mn) is approximately 54.94 g/mol, and the atomic mass of oxygen (O) is about 16.00 g/mol. Multiplying the atomic mass of oxygen by two, because there are two oxygen atoms in MnO\text{_}2, and then adding it to the atomic mass of manganese gives the molar mass of MnO\text{_}2.

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry that describes the behavior of ideal gases. It is stated as:
\ \( PV = nRT \ \).
Here, P is the pressure of the gas, V is its volume, n is the number of moles, R is the universal gas constant (0.0821 \text{ L·atm/(mol·K)}), and T is the temperature in Kelvin.

• At Standard Temperature and Pressure (STP), the conditions are set at 0°C (273.15 K) and 1 atm pressure.
• An important fact is that 1 mole of any gas at STP occupies a volume of 22.4 liters.

In this exercise, we used the fact that 1 mole of Cl\text{_}2 gas occupies 22.4 liters at STP to find that: \ \( \text{Moles of Cl}_2 = \frac{1.78 \text{ L}}{22.4 \text{ L/mol}} = 0.0795 \text{ mol} \ \).
We applied this principle to determine the required mass of MnO\text{_}2 necessary to produce a given volume of chlorine gas.