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Chapter 17: Problem 33

Chlorine gas was first prepared in 1774 by \(\mathrm{C}\). W. Scheele by oxidizing sodium chloride with manganese(IV) oxide. The reaction is $$ \begin{aligned} \mathrm{NaCl}(a q)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q)+\mathrm{MnO}_{2}(s) & \longrightarrow \\ \mathrm{Na}_{2} \mathrm{SO}_{4}(a q) &+\mathrm{MnCl}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{Cl}_{2}(g) \end{aligned} $$ Balance this equation.

Short Answer

Expert verified
The balanced equation is: \[ 2\mathrm{NaCl}(a q) +\mathrm{H}_{2} \mathrm{SO}_{4}(a q) +\mathrm{MnO}_{2}(s) \longrightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}(a q) +\mathrm{MnCl}_{2}(a q)+2\mathrm{H}_{2}\mathrm{O}(l)+\mathrm{Cl}_{2}(g) \]

Step by step solution

01

1. Write down the given unbalanced equation

The given unbalanced equation is: \[ \mathrm{NaCl}(a q)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q)+\mathrm{MnO}_{2}(s) \longrightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}(a q) +\mathrm{MnCl}_{2}(a q)+\mathrm{H}_{2}\mathrm{O}(l)+\mathrm{Cl}_{2}(g) \]
02

2. Balance chlorine atoms

Start by balancing Cl atoms since they are present in more than one molecule. We have 1 Cl atom on the left side and 3 Cl atoms on the right side (1 from Na2SO4 and 2 from MnCl2). To balance Cl atoms, add a coefficient of 2 in front of NaCl: \[ 2\mathrm{NaCl}(a q) +\mathrm{H}_{2} \mathrm{SO}_{4}(a q) +\mathrm{MnO}_{2}(s) \longrightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}(a q) +\mathrm{MnCl}_{2}(a q)+\mathrm{H}_{2}\mathrm{O}(l)+\mathrm{Cl}_{2}(g) \]
03

3. Balance sodium atoms

Next, balance Na atoms. We now have 2 Na atoms on both the left and right sides. So, no change is needed for Na atoms.
04

4. Balance manganese atoms

Now, balance Mn atoms. There is 1 Mn atom on both the left and right sides. So, no change is needed for Mn atoms.
05

5. Balance oxygen atoms

Balance O atoms. We have 6 O atoms on the left side (4 from H2SO4 and 2 from MnO2) and 5 O atoms on the right side (4 from Na2SO4 and 1 from H2O). To balance O atoms, add a coefficient of 2 in front of H2O: \[ 2\mathrm{NaCl}(a q) +\mathrm{H}_{2} \mathrm{SO}_{4}(a q) +\mathrm{MnO}_{2}(s) \longrightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}(a q) +\mathrm{MnCl}_{2}(a q)+2\mathrm{H}_{2}\mathrm{O}(l)+\mathrm{Cl}_{2}(g) \]
06

6. Balance hydrogen atoms

Lastly, balance H atoms. We now have 4 H atoms on the left side (2 from H2SO4) and 4 H atoms on the right side (4 from 2 H2O). So, no change is needed for H atoms.
07

7. Check if the equation is balanced

Verify if all elements are balanced in the equation: - Na: 2 atoms on both sides, - Cl: 2 atoms on both sides, - H: 4 atoms on both sides, - S: 1 atom on both sides, - Mn: 1 atom on both sides, - O: 6 atoms on both sides. Since all elements are balanced, the balanced equation is: \[ 2\mathrm{NaCl}(a q) +\mathrm{H}_{2} \mathrm{SO}_{4}(a q) +\mathrm{MnO}_{2}(s) \longrightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}(a q) +\mathrm{MnCl}_{2}(a q)+2\mathrm{H}_{2}\mathrm{O}(l)+\mathrm{Cl}_{2}(g) \]

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Most popular questions from this chapter

What is the maximum work that can be obtained from a hydrogen-oxygen fuel cell at standard conditions that produces \(1.00 \mathrm{kg}\) water at \(25^{\circ} \mathrm{C} ?\) Why do we say that this is the maximum work that can be obtained? What are the advantages and disadvantages in using fuel cells rather than the corresponding combustion reactions to produce electricity?

When magnesium metal is added to a beaker of \(\mathrm{HCl}(a q),\) a gas is produced. Knowing that magnesium is oxidized and that hydrogen is reduced, write the balanced equation for the reaction. How many electrons are transferred in the balanced equation? What quantity of useful work can be obtained when Mg is added directly to the beaker of HCl? How can you harness this reaction to do useful work?

Aluminum is produced commercially by the electrolysis of \(\mathrm{Al}_{2} \mathrm{O}_{3}\) in the presence of a molten salt. If a plant has a continuous capacity of 1.00 million \(A\), what mass of aluminum can be produced in \(2.00 \mathrm{h} ?\)

The measurement of \(\mathrm{pH}\) using a glass electrode obeys the Nernst equation. The typical response of a pH meter at \(25.00^{\circ} \mathrm{C}\) is given by the equation $$ \mathscr{E}_{\text {meas }}=\mathscr{E}_{\text {ref }}+0.05916 \mathrm{pH} $$ where \(\mathscr{E}_{\text {ref }}\) contains the potential of the reference electrode and all other potentials that arise in the cell that are not related to the hydrogen ion concentration. Assume that \(\mathscr{E}_{\mathrm{ref}}=0.250 \mathrm{V}\) and that \(\mathscr{E}_{\text {meas }}=0.480 \mathrm{V}\) a. What is the uncertainty in the values of \(\mathrm{pH}\) and \(\left[\mathrm{H}^{+}\right]\) if the uncertainty in the measured potential is \(\pm 1 \mathrm{mV}\) \((\pm 0.001 \mathrm{V}) ?\) b. To what precision must the potential be measured for the uncertainty in \(\mathrm{pH}\) to be \(\pm 0.02 \mathrm{pH}\) unit?

What volumes of \(\mathrm{H}_{2}(g)\) and \(\mathrm{O}_{2}(g)\) at \(\mathrm{STP}\) are produced from the electrolysis of water by a current of \(2.50 \mathrm{A}\) in \(15.0 \mathrm{min} ?\)
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