MediumChemistryHigh School (11–12)The p-Block ElementsNumerical

A sample of 5.43 g of mercury(II) chloride (271.49 g/mol) is reduced by tin(II) chloride (189.61…

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A sample of $5.43\ \text{g}$ of mercury(II) chloride ( $271.49\ \text{g/mol}$ ) is reduced by tin(II) chloride ( $189.61\ \text{g/mol}$ ). First white calomel forms: $2\text{HgCl}_{2} + \text{SnCl}_{2} \to \text{Hg}_{2}\text{Cl}_{2} + \text{SnCl}_{4}$ . Then more tin(II) chloride turns it into grey mercury: $\text{Hg}_{2}\text{Cl}_{2} + \text{SnCl}_{2} \to 2\text{Hg} + \text{SnCl}_{4}$ . Use $472.08\ \text{g/mol}$ for calomel and $200.59\ \text{g/mol}$ for mercury. Find the mass of calomel, the mass of tin(II) chloride for the first step, the mass of mercury at the end and the total mass of tin(II) chloride used.

Need a reminder? The key ideaShow

For a redox reaction, $E^\circ_{cell} = E^\circ_{ox} - E^\circ_{red}$ , $\Delta G^\circ = -nFE^\circ_{cell}$ with $F = 96.485\ \text{kJ/(V mol)}$ , and at 298 K, $\log K = \dfrac{nE^\circ_{cell}}{0.05916\ \text{V}}$ . A reaction goes if $E^\circ_{cell}$ is positive. Moles of reactant come from mass divided by molar mass, and the equation gives the ratios.

The worked solution

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