0.75 g of a sample of pyrolusite ore is treated with 1.89 g of oxalic acid crystals $(H_{2} C_{2} O_{4} . {2H}_{2} O)$ in acidic medium to analyze for its ${MnO}_{2}$ content. Following the reaction, the excess oxalic acid is titrated with 0.100 M ${KMnO}_{4}$ under acidic conditions, 30.00 mL being required. The percentage of Mn in the ore is______. (Atomic weights: Mn = 55 u)

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

answer is 55.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$M n O_{2}^{} + C_{2} O_{4}^{2 -} + 4 H^{+} \overset{}{\to} 2 M n^{2 +} + 2 C O_{2} + 2 H_{2} O$

$2 M n O_{4}^{-} + 5 C_{2} O_{4}^{2 -} + 16 H^{+} \overset{}{\to} 2 M n^{2 +} + 10 C O_{2} + 8 H_{2} O$

$2 \times n_{C_{2} O_{4}^{2 -}} (u n u s e d) = 5 \times n_{M n O_{4}^{-}} \Rightarrow 2 \times x = 5 \times \frac{0.1 \times 30.0}{1000} \Rightarrow x = 7.5 \times 10^{- 3} m o l$

$\begin{array}{l} n_{C_{2} O_{4}^{2 -}} u s e d) = \frac{1.89}{126} - 7.5 \times 10^{- 3} = 7.5 \times 10^{- 3} = n_{M n} (i n pyrolusite) \\ \Rightarrow % of Mn= \frac{7.5 \times 10^{- 3} \times 55}{0.75} \times 100 = 55 % \end{array}$