Question

A sample of impure tin of mass 0.538 g is dissolved in strong acid to give a solution of Sn2+. The solution is then titrated with a 0.0448 M solution of NO3−, which is reduced to NO(g). The equivalence point is reached upon the addition of 4.13×10−2 L of the NO3− solution.

Answer 1

The question is incomplete, here is a complete question.

A sample of impure tin of mass 0.538 g is dissolved in strong acid to give a solution of $Sn^{2+}$. The solution is then titrated with a 0.0448 M solution of $NO_3^-$, which is reduced to NO(g). The equivalence point is reached upon the addition of $4.13\times 10^{-2 }L$ of the $NO_3^-$ solution.

Find the percent by mass of tin in the original sample, assuming that it contains no other reducing agents.

Answer : The percent mass of tin in the original sample is 61.3 %

Solution :

First we have to calculate the moles of $NO_3^-$.

$\text{Moles of }NO_3^-=\text{Concentration of }NO_3^-\times \text{Volume of solution}$

$\text{Moles of }NO_3^-=0.0448M\times 4.13\times 10^{-2}L=1.85\times 10^{-3}mol$

Now we have to calculate the moles of $Sn^{4+}$

The balanced chemical reaction is,

$2NO_3^-(aq)+3Sn^{2+}(aq)+8H^+(aq)\rightarrow 2NO(g)+3Sn^{4+}(aq)+4H_2O(l)$

From the reaction, we conclude that

As, 2 mole of $NO_3^-$ react with 3 mole of $Sn^{2+}$

So, $1.85\times 10^{-3}$  moles of $NO_3^-$ react with $\frac{1.85\times 10^{-3}}{2}\times 3=2.78\times 10^{-3}$ moles of $Sn^{2+}$

Now we have to calculate the mass of $Sn^{2+}$

$\text{ Mass of }Sn^{2+}=\text{ Moles of }Sn^{2+}\times \text{ Molar mass of }Sn^{2+}$

Molar mass of tin = 118.71 g/mol

$\text{ Mass of }Sn^{2+}=(2.78\times 10^{-3}moles)\times (118.71g/mole)=0.330g$

Mass of $Sn^{2+}$ reacted = 0.330 g

Original mass of $Sn^{2+}$ = 0.538 g

Now we have to calculate the percent mass of tin in the original sample.

$\% \text{ mass of tin in the original sample}=\frac{\text{ Reacted mass of }Sn^{2+}}{\text{ Original mass of }Sn^{2+}}\times 100$

$\% \text{ mass of tin in the original sample}=\frac{0.330g}{0.538g}\times 100=61.3\%$

Therefore, the percent mass of tin in the original sample is 61.3 %