Question

CuFeS2 + 3 CuCl2 --> 4 CuCl + FeCl2 + 2 S How many mL of 0.075 M CuCl2 are needed to titrate 0.20 grams of CuFeS2?

Answer 1

Answer:

$V_{CuCl_2 }= 43.6mL$

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:

$3*n_{CuFeS_2}=n_{CuCl_2 }$

Thus, by introducing the mass and molar mass of the former and the volume and concentration of the latter, we write:

$3*0.20gCuFeS_2*\frac{1mol}{183.51g} =V_{CuCl_2 }*0.075\frac{mol}{L} \\\\V_{CuCl_2 }=\frac{0.00327mol}{0.075mol/L}*\frac{1000mL}{1L} \\\\V_{CuCl_2 }= 43.6mL$

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