Question

If you mix 200.0 mL of a 0.150 M solution of silver acetate with 2.500 g of copper metal you will get a single replacement reaction. What mass of Silver metal will you produce if the reaction has a 45.0% efficiency?

Answer 1

Answer: 2.11 g

Step-by-step explanation:

`$$Equation: $2 \mathrm{AgC}_(2) \mathrm{H}_(3) \mathrm{O}_(2)(a q)+\mathrm{Cu}(s) \longrightarrow \mathrm{Cu}\left(\mathrm{C}_(2) \mathrm{H}_(3) \mathrm{O}_(2)\right)_(2)(a q)+2 \mathrm{Ag}(s)$`

`$$No of moles of $\mathrm{Ag}$ produced from $\mathrm{Cu}=2.500 \mathrm{~g} * \frac{1 \mathrm{~mol} \mathrm{Cu}}{63.55 \mathrm{~g}} * \frac{2 \mathrm{~mol} \mathrm{Ag}}{1 \mathrm{~mol} \mathrm{Cu}}$ $=0.07868 \mathrm{~mol} \mathrm{Ag}$`

`$$No of moles of $\mathrm{Ag}$ produced from $\mathrm{AgC}_(2) \mathrm{H}_(3) \mathrm{O}_(2)=200.0 \mathrm{~mL} * \frac{1 \mathrm{~L}}{1000 \mathrm{~mL}} * \frac{0.150 \mathrm{~mol} \mathrm{AgC_(2 ) \mathrm { H } _ { 3 } \mathrm { O } _ { 2 }}}{1 \mathrm{~L}}$ $=0.0300 \mathrm{~mol} \mathrm{Ag}$`

`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 $\mathrm{Ag}=0.0300 \mathrm{~mol} \mathrm{Ag} * \frac{108 \mathrm{~g}}{1 \mathrm{~mol} \mathrm{Ag}}=3.24 \mathrm{~g}$\\Actual yield of silver $=65.0 \%(3.24 \mathrm{~g})$$$=(65.0)/(100)(3.24 \mathrm{~g})=2.106 \mathrm{~g} \approx 2.11 \mathrm{~g}$$`