Use Cu = 63.5, S = 32. A 0.700 g mixture of CuS and Cu2S reacts completely with 200 mL of 0.04 M acidified KMnO4 (MnO4- -> Mn^2+). In both cases the products are Cu^2+ and SO2. Find the mass percentage of Cu2S in the mixture (two decimals).

JEE Advanced levelNumerical answerDifficulty 3 of 3rx158

Use Cu = 63.5, S = 32. A $0.700 g$ mixture of $CuS$ and $Cu X_{2} S$ reacts completely with $200 mL$ of $0.04 M$ acidified $KMnO X_{4}$ ( $MnO X_{4} X^{-} Mn X^{2 +}$ ). In both cases the products are $Cu X^{2 +}$ and $SO X_{2}$ . Find the mass percentage of $Cu X_{2} S$ in the mixture (two decimals).

Numerical answer type, answer in %, tolerance 0.05.

n

-factors:

CuS

(

S X^{2 -} S X^{4 +}

)

= 6

;

Cu X_{2} S

= 2 + 6 = 8

. Let

x

= mol CuS,

y

= mol

Cu X_{2} S

. Equivalents:

6 x + 8 y = 0.200 \times 0.04 \times 5 = 0.04

. Mass:

95.5 x + 159 y = 0.700

. Solving gives

x = 0.0040

y = 0.0020

. Mass of

Cu X_{2} S

= 0.3180 g

, so

45.43%