301. In the neutral-medium half-reaction
\[
\mathrm{MnO_4^-}+2\mathrm{H_2O}+3e^-\rightarrow\mathrm{MnO_2}+x\mathrm{OH^-},
\]
the value of \(x\) is:
ⓐ. \(2\)
ⓑ. \(6\)
ⓒ. \(4\)
ⓓ. \(8\)
Correct Answer: \(4\)
Explanation: \( \textbf{Incomplete half-reaction:} \)
\[
\mathrm{MnO_4^-}+2\mathrm{H_2O}+3e^-\rightarrow\mathrm{MnO_2}+x\mathrm{OH^-}
\]
\( \textbf{Hydrogen balance:} \) The left side contains \(4\) hydrogen atoms, so the right side must contain \(4\mathrm{OH^-}\).
\[
x=4
\]
\( \textbf{Oxygen check:} \) The left side contains \(4+2=6\) oxygen atoms.
\[
2+4=6
\]
The oxygen atoms are therefore balanced when \(x=4\).
\( \textbf{Charge check:} \) The total charge on the left is \(-1-3=-4\), while \(4\mathrm{OH^-}\) gives \(-4\) on the right.
\( \textbf{Completed half-reaction:} \)
\[
\mathrm{MnO_4^-}+2\mathrm{H_2O}+3e^-\rightarrow\mathrm{MnO_2}+4\mathrm{OH^-}
\]
\( \textbf{Final answer:} \) The missing coefficient is \(4\).
302. For one mole of permanganate, the order of decreasing oxidising capacity based on the number of electrons accepted is:
ⓐ. strongly alkaline \(\gt\) neutral \(\gt\) acidic
ⓑ. acidic \(\gt\) neutral \(\gt\) strongly alkaline
ⓒ. neutral \(\gt\) strongly alkaline \(\gt\) acidic
ⓓ. acidic \(\gt\) strongly alkaline \(\gt\) neutral
Correct Answer: acidic \(\gt\) neutral \(\gt\) strongly alkaline
Explanation: One mole of permanganate accepts five moles of electrons when reduced to \(\mathrm{Mn^{2+}}\) in acidic medium. It accepts three moles of electrons when reduced to \(\mathrm{MnO_2}\) in neutral or weakly alkaline medium. It accepts only one mole of electrons when reduced to manganate in strongly alkaline medium. On an electron-capacity basis, the order is therefore \(5\gt3\gt1\). This comparison refers to the specified reduction products. The medium must always be stated before assigning the electron capacity.
303. A reaction is performed in a strongly alkaline solution, but the product equation is written with \(\mathrm{Mn^{2+}}\) as the permanganate reduction product. The most likely diagnosis is:
ⓐ. the equation is correct because \(\mathrm{Mn^{2+}}\) forms in every medium
ⓑ. manganate, not \(\mathrm{Mn^{2+}}\), is expected in strong alkali
ⓒ. the manganese oxidation state should remain \(+7\)
ⓓ. permanganate cannot undergo reduction in alkaline solution
Correct Answer: manganate, not \(\mathrm{Mn^{2+}}\), is expected in strong alkali
Explanation: Strongly alkaline conditions favour reduction of permanganate to \(\mathrm{MnO_4^{2-}}\). Manganese then changes from \(+7\) to \(+6\) through a one-electron gain. Formation of \(\mathrm{Mn^{2+}}\) is characteristic of acidic reduction and requires five electrons. Using that product in a strongly alkaline equation gives an incorrect \(n\)-factor and incorrect stoichiometric coefficients. Permanganate can certainly undergo reduction in alkaline medium. The product must be chosen from the stated reaction conditions before balancing the equation.
304. A process requires permanganate to be reduced without forming either dissolved \(\mathrm{Mn^{2+}}\) or a solid \(\mathrm{MnO_2}\) precipitate. The most suitable condition and product are:
ⓐ. strong alkali, giving soluble \(\mathrm{MnO_4^{2-}}\)
ⓑ. acidic medium, producing soluble \(\mathrm{Mn^{2+}}\)
ⓒ. neutral medium, producing soluble \(\mathrm{MnO_2}\)
ⓓ. weakly alkaline medium, producing manganese metal
Correct Answer: strong alkali, giving soluble \(\mathrm{MnO_4^{2-}}\)
Explanation: Acidic reduction would produce dissolved \(\mathrm{Mn^{2+}}\), which the process excludes. Neutral or weakly alkaline conditions commonly produce solid \(\mathrm{MnO_2}\), which would create a precipitate. Strongly alkaline medium instead favours formation of the green manganate ion. Manganate remains an oxoanion in solution and involves a one-electron reduction of permanganate. Manganese changes only from \(+7\) to \(+6\). The strongly alkaline route therefore satisfies both restrictions in the question.
305. Acidified permanganate is added to a solution containing \(\mathrm{Fe^{2+}}\). How do the oxidation states change?
ⓐ. \(\mathrm{Fe^{2+}}\) becomes \(\mathrm{Fe^{3+}}\), while permanganate becomes \(\mathrm{Mn^{2+}}\)
ⓑ. \(\mathrm{Fe^{2+}}\) is reduced to iron metal, while permanganate is oxidised to manganate
ⓒ. Both \(\mathrm{Fe^{2+}}\) and permanganate are oxidised
ⓓ. \(\mathrm{Fe^{3+}}\) is reduced to \(\mathrm{Fe^{2+}}\), while \(\mathrm{Mn^{2+}}\) forms permanganate
Correct Answer: \(\mathrm{Fe^{2+}}\) becomes \(\mathrm{Fe^{3+}}\), while permanganate becomes \(\mathrm{Mn^{2+}}\)
Explanation: Iron(II) loses one electron and forms iron(III). Permanganate accepts electrons in acidic medium and is reduced from manganese(VII) to manganese(II). The \(\mathrm{Fe^{2+}}\) ion therefore acts as the reducing agent. Permanganate acts as the oxidising agent. The purple colour of permanganate disappears as the reaction proceeds. Both oxidation and reduction occur simultaneously through electron transfer.
306. Acidified permanganate oxidises \(\mathrm{Fe^{2+}}\). The balanced ionic equation is:
ⓐ. \(\mathrm{MnO_4^-}+4\mathrm{H^+}+3\mathrm{Fe^{2+}}\rightarrow\mathrm{MnO_2}+2\mathrm{H_2O}+3\mathrm{Fe^{3+}}\)
ⓑ. \(\mathrm{MnO_4^-}+8\mathrm{H^+}+5\mathrm{Fe^{3+}}\rightarrow\mathrm{Mn^{2+}}+4\mathrm{H_2O}+5\mathrm{Fe^{2+}}\)
ⓒ. \(\mathrm{MnO_4^-}+8\mathrm{H^+}+5\mathrm{Fe^{2+}}\rightarrow\mathrm{Mn^{2+}}+4\mathrm{H_2O}+5\mathrm{Fe^{3+}}\)
ⓓ. \(\mathrm{MnO_4^-}+5\mathrm{H^+}+\mathrm{Fe^{2+}}\rightarrow\mathrm{Mn^{2+}}+5\mathrm{H_2O}+\mathrm{Fe^{3+}}\)
Correct Answer: \(\mathrm{MnO_4^-}+8\mathrm{H^+}+5\mathrm{Fe^{2+}}\rightarrow\mathrm{Mn^{2+}}+4\mathrm{H_2O}+5\mathrm{Fe^{3+}}\)
Explanation: Permanganate accepts five electrons when reduced to \(\mathrm{Mn^{2+}}\) in acidic medium. Each \(\mathrm{Fe^{2+}}\) ion loses one electron while forming \(\mathrm{Fe^{3+}}\). Five iron(II) ions are therefore required for one permanganate ion. Eight hydrogen ions and four water molecules balance hydrogen and oxygen. The total charge on the reactant side is \(-1+8+10=+17\). The product-side charge is \(+2+15=+17\), confirming complete charge balance.
307. Complete reaction of \(0.025\,mol\) of \(\mathrm{MnO_4^-}\) with \(\mathrm{Fe^{2+}}\) in acidic medium requires:
ⓐ. \(0.050\,mol\) of \(\mathrm{Fe^{2+}}\)
ⓑ. \(0.125\,mol\) of \(\mathrm{Fe^{2+}}\)
ⓒ. \(0.025\,mol\) of \(\mathrm{Fe^{2+}}\)
ⓓ. \(0.250\,mol\) of \(\mathrm{Fe^{2+}}\)
Correct Answer: \(0.125\,mol\) of \(\mathrm{Fe^{2+}}\)
Explanation: \( \textbf{Balanced mole relation:} \)
\[
1\,mol\ \mathrm{MnO_4^-}:5\,mol\ \mathrm{Fe^{2+}}
\]
\( \textbf{Permanganate amount:} \)
\[
n(\mathrm{MnO_4^-})=0.025\,mol
\]
\( \textbf{Required iron(II) amount:} \)
\[
n(\mathrm{Fe^{2+}})=5\times0.025
\]
\[
n(\mathrm{Fe^{2+}})=0.125\,mol
\]
\( \textbf{Electron check:} \)
\[
0.125\,mol\ \mathrm{Fe^{2+}}\text{ releases }0.125\,mol\ e^-
\]
\( \textbf{Permanganate electron requirement:} \)
\[
0.025\times5=0.125\,mol\ e^-
\]
\( \textbf{Final answer:} \) The reaction requires \(0.125\,mol\) of \(\mathrm{Fe^{2+}}\).
308. For oxidation of iodide by acidified permanganate, the balanced ionic equation is:
ⓐ. \(\mathrm{MnO_4^-}+8\mathrm{H^+}+5\mathrm{I^-}\rightarrow\mathrm{Mn^{2+}}+4\mathrm{H_2O}+5\mathrm{I_2}\)
ⓑ. \(\mathrm{2MnO_4^-}+8\mathrm{H^+}+10\mathrm{I^-}\rightarrow2\mathrm{Mn^{2+}}+4\mathrm{H_2O}+5\mathrm{I_2}\)
ⓒ. \(\mathrm{2MnO_4^-}+16\mathrm{H^+}+5\mathrm{I^-}\rightarrow2\mathrm{Mn^{2+}}+8\mathrm{H_2O}+5\mathrm{I_2}\)
ⓓ. \(\mathrm{2MnO_4^-}+16\mathrm{H^+}+10\mathrm{I^-}\rightarrow2\mathrm{Mn^{2+}}+8\mathrm{H_2O}+5\mathrm{I_2}\)
Correct Answer: \(\mathrm{2MnO_4^-}+16\mathrm{H^+}+10\mathrm{I^-}\rightarrow2\mathrm{Mn^{2+}}+8\mathrm{H_2O}+5\mathrm{I_2}\)
Explanation: Two permanganate ions accept a total of ten electrons in acidic medium. Ten iodide ions release ten electrons while forming five iodine molecules. Sixteen hydrogen ions and eight water molecules balance hydrogen and oxygen. The total charge on the reactant side is \(-2+16-10=+4\). Two manganese(II) ions on the product side also give charge \(+4\). The equation therefore conserves atoms, electrons, and total charge.
309. Acidified potassium permanganate is added to excess potassium iodide, followed by starch solution. Which observation is expected?
ⓐ. a green manganate solution forms without production of molecular iodine
ⓑ. brown \(\mathrm{MnO_2}\) precipitates while iodide remains unchanged
ⓒ. purple colour disappears, and liberated iodine turns starch blue-black
ⓓ. the solution remains purple because iodide cannot reduce permanganate
Correct Answer: purple colour disappears, and liberated iodine turns starch blue-black
Explanation: Iodide acts as the reducing agent and is oxidised to iodine. Acidified permanganate accepts electrons and is reduced to \(\mathrm{Mn^{2+}}\). Consumption of permanganate removes its intense purple colour. The iodine formed produces a brown solution before starch is added. Iodine forms a characteristic blue-black complex with starch. The observations therefore provide evidence for both permanganate reduction and iodide oxidation.
310. A graph plots moles of \(\mathrm{I_2}\) formed on the vertical axis against moles of \(\mathrm{MnO_4^-}\) consumed in the oxidation of excess iodide by acidified permanganate. The slope of the straight line is:
ⓐ. \(2.5\)
ⓑ. \(5.0\)
ⓒ. \(0.40\)
ⓓ. \(1.0\)
Correct Answer: \(2.5\)
Explanation: \( \textbf{Balanced stoichiometric relation:} \)
\[
2\mathrm{MnO_4^-}+16\mathrm{H^+}+10\mathrm{I^-}\rightarrow2\mathrm{Mn^{2+}}+8\mathrm{H_2O}+5\mathrm{I_2}
\]
\( \textbf{Graph variables:} \) The vertical axis is \(n(\mathrm{I_2})\), and the horizontal axis is \(n(\mathrm{MnO_4^-})\).
\( \textbf{Slope relation:} \)
\[
\text{slope}=\frac{\Delta n(\mathrm{I_2})}{\Delta n(\mathrm{MnO_4^-})}
\]
\( \textbf{Use the coefficients:} \)
\[
\text{slope}=\frac{5}{2}=2.5
\]
\( \textbf{Final answer:} \) The graph is a straight line through the origin with slope \(2.5\).
311. A learner claims that one mole of acidic permanganate oxidises five moles of iodine molecules because permanganate accepts five electrons. The best correction is:
ⓐ. One mole of permanganate oxidises \(10\) moles of iodide to \(5\) moles of iodine
ⓑ. One mole of permanganate oxidises \(5\) moles of iodide to \(2.5\) moles of iodine
ⓒ. One mole of permanganate oxidises \(5\) moles of iodine to iodide
ⓓ. One mole of permanganate oxidises \(1\) mole of iodide to \(0.5\) mole of iodine
Correct Answer: One mole of permanganate oxidises \(5\) moles of iodide to \(2.5\) moles of iodine
Explanation: One iodide ion loses one electron during oxidation. One permanganate ion accepts five electrons in acidic medium. It therefore oxidises five iodide ions. Two iodide ions combine to form one iodine molecule. Five iodide ions consequently produce \(5/2=2.5\) iodine molecules on a molar basis. The learner has confused the number of iodide ions oxidised with the number of iodine molecules formed.
312. The oxidation half-reaction for oxalate ions is:
ⓐ. \(\mathrm{C_2O_4^{2-}}\rightarrow2\mathrm{CO_2}+2e^-\)
ⓑ. \(\mathrm{C_2O_4^{2-}}+2e^-\rightarrow2\mathrm{CO_2}\)
ⓒ. \(\mathrm{C_2O_4^{2-}}\rightarrow\mathrm{CO_2}+2\mathrm{O_2}+2e^-\)
ⓓ. \(\mathrm{2CO_2}+2e^-\rightarrow\mathrm{C_2O_4^{2-}}\)
Correct Answer: \(\mathrm{C_2O_4^{2-}}\rightarrow2\mathrm{CO_2}+2e^-\)
Explanation: Oxalate contains two carbon atoms, and both appear in the two carbon dioxide molecules. The four oxygen atoms are also conserved. The charge on the reactant side is \(-2\). The two electrons on the product side provide the same total charge of \(-2\). Carbon has oxidation state \(+3\) in oxalate and \(+4\) in carbon dioxide. The two carbon atoms therefore lose one electron each, giving two electrons altogether.
313. For oxidation of oxalate by acidified permanganate, the balanced ionic equation is:
ⓐ. \(\mathrm{MnO_4^-}+8\mathrm{H^+}+5\mathrm{C_2O_4^{2-}}\rightarrow\mathrm{Mn^{2+}}+4\mathrm{H_2O}+10\mathrm{CO_2}\)
ⓑ. \(\mathrm{2MnO_4^-}+8\mathrm{H^+}+5\mathrm{C_2O_4^{2-}}\rightarrow2\mathrm{Mn^{2+}}+4\mathrm{H_2O}+10\mathrm{CO_2}\)
ⓒ. \(\mathrm{2MnO_4^-}+16\mathrm{H^+}+5\mathrm{C_2O_4^{2-}}\rightarrow2\mathrm{Mn^{2+}}+8\mathrm{H_2O}+10\mathrm{CO_2}\)
ⓓ. \(\mathrm{5MnO_4^-}+16\mathrm{H^+}+2\mathrm{C_2O_4^{2-}}\rightarrow5\mathrm{Mn^{2+}}+8\mathrm{H_2O}+4\mathrm{CO_2}\)
Correct Answer: \(\mathrm{2MnO_4^-}+16\mathrm{H^+}+5\mathrm{C_2O_4^{2-}}\rightarrow2\mathrm{Mn^{2+}}+8\mathrm{H_2O}+10\mathrm{CO_2}\)
Explanation: Two permanganate ions require ten electrons for reduction to two manganese(II) ions. Each oxalate ion releases two electrons when oxidised to carbon dioxide. Five oxalate ions therefore supply the required ten electrons. Sixteen hydrogen ions and eight water molecules balance hydrogen and oxygen. The ten carbon atoms in five oxalate ions form ten carbon dioxide molecules. The total charge on both sides is \(+4\), confirming that the equation is balanced.
314. In a practical titration of acidified permanganate with oxalate, the reaction mixture is warmed initially because:
ⓐ. warming accelerates the initially slow room-temperature reaction
ⓑ. heating changes permanganate into a reducing agent
ⓒ. oxalate reacts only after water has completely evaporated
ⓓ. warming changes the required permanganate-to-oxalate ratio from \(2:5\) to \(1:1\)
Correct Answer: warming accelerates the initially slow room-temperature reaction
Explanation: The permanganate-oxalate reaction has a significant kinetic barrier at ordinary room temperature. Warming increases the fraction of collisions with sufficient energy to react. The reaction then proceeds at a rate suitable for volumetric analysis. The stoichiometric ratio remains \(2:5\) because heating does not change the balanced redox equation. Excessive boiling is unnecessary and may create experimental errors. Moderate warming is used to improve reaction rate rather than to alter the chemical products.
315. Assertion: The oxidation of oxalate by acidified permanganate often accelerates after the reaction has started.
Reason: The \(\mathrm{Mn^{2+}}\) formed during the reaction can catalyse the same reaction.
ⓐ. Assertion is true, but Reason is false
ⓑ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓒ. Assertion is false, but Reason is true
ⓓ. Both Assertion and Reason are true, and Reason explains Assertion
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: At the beginning, very little manganese(II) is present and the reaction may be relatively slow. Permanganate reduction gradually produces \(\mathrm{Mn^{2+}}\). These manganese(II) ions facilitate a faster reaction pathway. As their concentration rises, the reaction rate increases. A product that catalyses the reaction producing it gives autocatalytic behaviour. The Reason therefore explains the acceleration described in the Assertion.
316. A graph of reaction rate against time is drawn for the warmed reaction between acidified permanganate and oxalate. Which qualitative behaviour is most reasonable?
ⓐ. the rate remains zero because oxalate cannot react with permanganate
ⓑ. the rate decreases uniformly because no \(\mathrm{Mn^{2+}}\) catalyst forms
ⓒ. the rate rises indefinitely as \(\mathrm{Mn^{2+}}\) accumulates
ⓓ. rate rises as \(\mathrm{Mn^{2+}}\) forms, then falls as reactants deplete
Correct Answer: rate rises as \(\mathrm{Mn^{2+}}\) forms, then falls as reactants deplete
Explanation: The reaction begins relatively slowly because the initial manganese(II) concentration is small. As permanganate is reduced, \(\mathrm{Mn^{2+}}\) forms and catalyses further reaction. The rate therefore increases during the early and middle stages. Eventually, the concentrations of permanganate and oxalate decrease substantially. Reactant depletion then causes the rate to fall. The resulting pattern reflects competition between autocatalyst formation and consumption of reactants.
317. Complete reaction of \(0.020\,mol\) of permanganate with oxalate produces carbon dioxide. If the molar gas volume under the stated conditions is \(22.4\,L\,mol^{-1}\), the volume of \(\mathrm{CO_2}\) formed is:
ⓐ. \(1.12\,L\)
ⓑ. \(2.24\,L\)
ⓒ. \(4.48\,L\)
ⓓ. \(0.448\,L\)
Correct Answer: \(2.24\,L\)
Explanation: \( \textbf{Balanced product relation:} \)
\[
2\,mol\ \mathrm{MnO_4^-}\rightarrow10\,mol\ \mathrm{CO_2}
\]
\( \textbf{Simplified relation:} \)
\[
1\,mol\ \mathrm{MnO_4^-}\rightarrow5\,mol\ \mathrm{CO_2}
\]
\( \textbf{Permanganate amount:} \)
\[
n(\mathrm{MnO_4^-})=0.020\,mol
\]
\( \textbf{Carbon dioxide amount:} \)
\[
n(\mathrm{CO_2})=5\times0.020=0.100\,mol
\]
\( \textbf{Gas-volume relation:} \)
\[
V=nV_m
\]
\( \textbf{Substitution:} \)
\[
V=0.100\times22.4
\]
\[
V=2.24\,L
\]
\( \textbf{Final answer:} \) The reaction produces \(2.24\,L\) of carbon dioxide.
318. A mixture contains \(0.040\,mol\) of permanganate and \(0.080\,mol\) of oxalate in sufficient acid. What remains after the reaction goes to completion?
ⓐ. Oxalate is limiting, but all \(0.040\,mol\) of permanganate reacts
ⓑ. Permanganate is limiting, and \(0.020\,mol\) of oxalate remains
ⓒ. Both reactants are consumed completely because their mole ratio is \(1:2\)
ⓓ. Oxalate is limiting, and \(0.008\,mol\) of permanganate remains
Correct Answer: Oxalate is limiting, and \(0.008\,mol\) of permanganate remains
Explanation: \( \textbf{Balanced reactant ratio:} \)
\[
2\,mol\ \mathrm{MnO_4^-}:5\,mol\ \mathrm{C_2O_4^{2-}}
\]
\( \textbf{Oxalate required for all permanganate:} \)
\[
0.040\times\frac{5}{2}=0.100\,mol
\]
\( \textbf{Oxalate actually available:} \)
\[
0.080\,mol
\]
\( \textbf{Limiting reactant:} \) Since \(0.080\,mol\lt0.100\,mol\), oxalate is limiting.
\( \textbf{Permanganate that reacts:} \)
\[
n(\mathrm{MnO_4^-})=0.080\times\frac{2}{5}
\]
\[
n(\mathrm{MnO_4^-})=0.032\,mol
\]
\( \textbf{Permanganate remaining:} \)
\[
0.040-0.032=0.008\,mol
\]
\( \textbf{Final answer:} \) Oxalate is limiting, and \(0.008\,mol\) of permanganate remains unreacted.
319. Acidified potassium permanganate is passed through a solution containing hydrogen sulphide. Which observation is most consistent with the redox reaction?
ⓐ. The solution becomes green without formation of any solid
ⓑ. A brown \(\mathrm{MnO_2}\) precipitate forms while hydrogen sulphide remains unchanged
ⓒ. The purple colour becomes more intense and sulphur dissolves completely
ⓓ. The purple colour is discharged and elemental sulphur is formed
Correct Answer: The purple colour is discharged and elemental sulphur is formed
Explanation: Acidified permanganate acts as a strong oxidising agent. It accepts electrons and is reduced from manganese(VII) to \(\mathrm{Mn^{2+}}\). Hydrogen sulphide acts as the reducing agent and is oxidised from sulphur oxidation state \(-2\) to elemental sulphur with oxidation state \(0\). Consumption of permanganate removes its intense purple colour. The elemental sulphur appears as a pale yellow solid or turbidity. Formation of brown \(\mathrm{MnO_2}\) would instead indicate permanganate reduction under neutral or weakly alkaline conditions.
320. The ionic equation for oxidation of hydrogen sulphide by acidified permanganate is written with one missing coefficient:
\[
2\mathrm{MnO_4^-}+x\mathrm{H^+}+5\mathrm{H_2S}\rightarrow2\mathrm{Mn^{2+}}+8\mathrm{H_2O}+5\mathrm{S}.
\]
The value of \(x\) is:
ⓐ. \(4\)
ⓑ. \(10\)
ⓒ. \(6\)
ⓓ. \(16\)
Correct Answer: \(6\)
Explanation: \( \textbf{Incomplete equation:} \)
\[
2\mathrm{MnO_4^-}+x\mathrm{H^+}+5\mathrm{H_2S}\rightarrow2\mathrm{Mn^{2+}}+8\mathrm{H_2O}+5\mathrm{S}
\]
\( \textbf{Reactant-side charge:} \)
\[
-2+x
\]
\( \textbf{Product-side charge:} \)
\[
2(+2)=+4
\]
\( \textbf{Charge balance:} \)
\[
-2+x=+4
\]
\[
x=6
\]
\( \textbf{Hydrogen check:} \) The reactants then contain \(6+5(2)=16\) hydrogen atoms, matching the \(16\) hydrogen atoms in \(8\mathrm{H_2O}\).
\( \textbf{Atom check:} \) Oxygen and sulphur are already balanced.
\( \textbf{Final answer:} \) The missing coefficient is \(x=6\).
