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Answer: 2.11 g
Explanation:
[tex]$Equation: $2 \mathrm{AgC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(a q)+\mathrm{Cu}(s) \longrightarrow \mathrm{Cu}\left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)_{2}(a q)+2 \mathrm{Ag}(s)$[/tex]
[tex]$No of moles of $\mathrm{Ag}$ produced from $\mathrm{Cu}=2.500 \mathrm{~g} \times \frac{1 \mathrm{~mol} \mathrm{Cu}}{63.55 \mathrm{~g}} \times \frac{2 \mathrm{~mol} \mathrm{Ag}}{1 \mathrm{~mol} \mathrm{Cu}}$ $=0.07868 \mathrm{~mol} \mathrm{Ag}$[/tex]
[tex]$No of moles of $\mathrm{Ag}$ produced from $\mathrm{AgC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}=200.0 \mathrm{~mL} \times \frac{1 \mathrm{~L}}{1000 \mathrm{~mL}} \times \frac{0.150 \mathrm{~mol} \mathrm{AgC_{2 } \mathrm { H } _ { 3 } \mathrm { O } _ { 2 }}}{1 \mathrm{~L}}$ $=0.0300 \mathrm{~mol} \mathrm{Ag}$[/tex]
[tex]Since $\mathrm{AgC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}$ limits the production of $\mathrm{Ag}, \mathrm{AgC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}$ acts as the limiting reactant \\\\No of moles of Ag formed $=0.0300 \mathrm{~mol}$\\Theoretical yield of $\mathrm{Ag}=0.0300 \mathrm{~mol} \mathrm{Ag} \times \frac{108 \mathrm{~g}}{1 \mathrm{~mol} \mathrm{Ag}}=3.24 \mathrm{~g}$\\Actual yield of silver $=65.0 \%(3.24 \mathrm{~g})$$=\frac{65.0}{100}(3.24 \mathrm{~g})=2.106 \mathrm{~g} \approx 2.11 \mathrm{~g}$[/tex]