The ionic equation shows the displacement of lead from its salt solution.
\(Al(s)+Pb^{2+}(aq)\rightarrow Al^{3+}(aq)+Pb(s)\\ \Delta H=-150.7\,kJ\,mol^{-1}\)
The reaction displaced 20.7 g of lead solid.
Calculate the amount of heat released from the reaction.
[Relative atomic mass: \(Pb=207\)]
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Answer: \(\text{15 070 J}\)
\(\text{Number of mole Pb}\\ =\dfrac{\text{Mass of Pb}}{\text{Molar mass of Pb}}\\ \,\\=\dfrac{20.7\,g}{207\,g\,mol^{-1}}\\ \,\\=0.1\,mol\)
\(\begin{aligned} \dfrac{\text{Heat released}}{0.1\,mol}&=\dfrac{150\,700\,J}{1\,mol} \end{aligned}\)
\(\text{Heat released}\\ =\dfrac{150\,700\,J}{1\,mol} \times 0.1\,mol\\ \,\\=\text{15 070 J}\)
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