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

`V_(CuCl_2 )= 43.6mL`

Step-by-step 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*(1mol)/(183.51g) =V_(CuCl_2 )*0.075(mol)/(L) \\\\V_(CuCl_2 )=(0.00327mol)/(0.075mol/L)*(1000mL)/(1L) \\\\V_(CuCl_2 )= 43.6mL`

Regards!