Question

A 2.0 g sample containing MnO2 is treated with HCl liberating Cl2. The Cl2 gas is passed into a solution of KI and 60.0 mL of 0.1 M Na2S2O3 is required to titrate the liberated iodine. The percentage of MnO2 in the sample is ______. (Nearest integer)
[Atomic masses (in u) Mn = 55; Cl = 35.5; O = 16, I = 127, Na = 23, K = 39, S = 32]

Answer 1

$MnO_2 \+ 4HCl → MnCl_2 \+ 2H_2O + Cl_2$
$Cl_2 + Kl + CL^- + I_2\ 2Na_2S_2O_3 → Na_2S_4O3$
Equivalent of MnO2 = HCl = Cl2 = I2 = Na2S2O3
2 × number of moles of MnO2 = 1 × number of moles of Na2S2O3
Moles of MnO2
$=\frac {60 × 0.1 × 10-3}{2}$
= $3×10^{–3}$ mole
Mass of MnO2 = 0.261 g
% of MnO2
$= \frac {0.261}{2}× 100$
$≃$ $13$So, the answer is $13$.