A $$2.0$$ g sample containing $$MnO_2$$ is treated with HCl liberating $$Cl_2$$. The $$Cl_2$$ gas is passed into a solution of KI and $$60.0$$ mL of $$0.1$$ M $$Na_2S_2O_3$$ is required to titrate the liberated iodine. The percentage of $$MnO_2$$ 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]

We need to find the percentage of $$MnO_2$$ in a 2.0 g sample, given that the liberated $$Cl_2$$ reacts with KI and the liberated $$I_2$$ requires 60.0 mL of 0.1 M $$Na_2S_2O_3$$.

When $$MnO_2$$ reacts with HCl according to $$MnO_2 + 4HCl \rightarrow MnCl_2 + Cl_2 + 2H_2O$$ one mole of $$MnO_2$$ produces one mole of $$Cl_2$$. The generated $$Cl_2$$ then oxidizes KI in the reaction $$Cl_2 + 2KI \rightarrow 2KCl + I_2$$ liberating one mole of $$I_2$$ per mole of $$Cl_2$$. The liberated $$I_2$$ is titrated with $$Na_2S_2O_3$$ according to $$I_2 + 2Na_2S_2O_3 \rightarrow Na_2S_4O_6 + 2NaI$$ where each mole of $$I_2$$ consumes two moles of $$Na_2S_2O_3$$.

The moles of $$Na_2S_2O_3$$ used in the titration are given by $$\text{Moles of } Na_2S_2O_3 = M \times V = 0.1 \times \frac{60}{1000} = 6 \times 10^{-3} \text{ mol}$$. Therefore, $$\text{Moles of } I_2 = \frac{\text{Moles of } Na_2S_2O_3}{2} = \frac{6 \times 10^{-3}}{2} = 3 \times 10^{-3} \text{ mol}$$. Since one mole of $$Cl_2$$ yields one mole of $$I_2$$, the moles of $$Cl_2$$ are also $$3 \times 10^{-3}$$ mol, which means the sample contained $$3 \times 10^{-3}$$ mol of $$MnO_2$$.

Using the molar mass of $$MnO_2$$ (55 + 2(16) = 87 g/mol), the mass of $$MnO_2$$ in the sample is $$3 \times 10^{-3} \times 87 = 0.261 \text{ g}$$. Hence, the percentage of $$MnO_2$$ in the 2.0 g sample is $$\frac{0.261}{2.0} \times 100 = 13.05\%$$, which rounds to 13%.

The correct answer is 13%.