Answer:
[tex]m_{AgCl}=7.15gAgCl\\m_{Ca\ Acet}=3.15gCa\ Acet\\[/tex]
Explanation:
Hello,
In this case, with the given information, we can identify the limiting reactant and compute the theoretical yield for the undergoing chemical reaction:
[tex]2CH_3COOAg+CaCl_2\rightarrow (CH_3COO)_2Ca+2AgCl[/tex]
Thus, with the given concentrations and volumes we compute the available moles of silver acetate:
[tex]n_{Ag\ Acet}=2.0mol/L*0.025mL=0.05mol[/tex]
Then, the moles of silver acetate that are consumed by 35 mL of 1.0 M calcium chloride:
[tex]n=0.035mL*1.0molCaCl_2/L*\frac{2mol Ag\ Acet}{1molCaCl_2} =0.07mol Ag\ Acet[/tex]
Therefore, since there are less available moles, it is the limiting reactant, for that reason, the theoretical yields of both calcium acetate and silver acetate are:
[tex]m_{AgCl}=0.05mol Ag\ Acet*\frac{2molAgCl}{2mol Ag\ Acet} *\frac{143gAgCl}{1molAgCl} \\\\m_{AgCl}=7.15gAgCl\\\\m_{Ca\ Acet}=0.05mol Ag\ Acet*\frac{1molmol Ca\ Acet}{2molmol Ag\ Acet} *\frac{126gmol Ca\ Acet}{1mol Ca\ Acet} \\\\m_{Ca\ Acet}=3.15gCa\ Acet[/tex]
Best regards.
