Question

25 ml of 2.0 M silver acetate reacts with grams 35 ml of 1.0 M Calcium chloride

Answer 1

Answer:

$m_{AgCl}=7.15gAgCl\\m_{Ca\ Acet}=3.15gCa\ Acet$

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:

$2CH_3COOAg+CaCl_2\rightarrow (CH_3COO)_2Ca+2AgCl$

Thus, with the given concentrations and volumes we compute the available moles of silver acetate:

$n_{Ag\ Acet}=2.0mol/L*0.025mL=0.05mol$

Then, the moles of silver acetate that are consumed by 35 mL of 1.0 M calcium chloride:

$n=0.035mL*1.0molCaCl_2/L*\frac{2mol Ag\ Acet}{1molCaCl_2} =0.07mol Ag\ Acet$

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:

$m_{AgCl}=0.05mol Ag\ Acet*\frac{2molAgCl}{2mol Ag\ Acet} *\frac{143gAgCl}{1molAgCl} \\\\m_{AgCl}=7.15gAgCl\\\\m_{Ca\ Acet}=0.05mol Ag\ Acet*\frac{1molmol Ca\ Acet}{2molmol Ag\ Acet} *\frac{126gmol Ca\ Acet}{1mol Ca\ Acet} \\\\m_{Ca\ Acet}=3.15gCa\ Acet$

Best regards.