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?

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:

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

Regards!

i believe it is c.) 1575 j.

answeridentity;