Answer:
[tex]V_{CuCl_2 }= 43.6mL[/tex]
Explanation:
Hello there!
In this case, according to this titration problem, it is possible to note there is a 1:3 mole ratio of CuFeS2 to CuCl2; it means that we can use the following equation:
[tex]3*n_{CuFeS_2}=n_{CuCl_2 }[/tex]
Thus, by introducing the mass and molar mass of the former and the volume and concentration of the latter, we write:
[tex]3*0.20gCuFeS_2*\frac{1mol}{183.51g} =V_{CuCl_2 }*0.075\frac{mol}{L} \\\\V_{CuCl_2 }=\frac{0.00327mol}{0.075mol/L}*\frac{1000mL}{1L} \\\\V_{CuCl_2 }= 43.6mL[/tex]
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