Question

Find the percent by mass of tin in the original sample, assuming that it contains no other reducing agents. A sample of impure tin of mass 0.528 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 4.03×10−2 L of the NO3− solution.

Answer 1

Explanation:

The balanced equation for the given reaction is as follows.

        $2NO^{-}_{3}(aq) + 3Sn^{2+}(aq) + 8H^{+} \rightarrow 2NO(g) + 3Sn^{4+} + 4H_{2}O$

Number of moles of $NO^{-}_{3}$ consumed will be calculated as follows.

    No. of moles = $Molarity \times {\text{volume in L}}$

                          = $0.0448 M \times 4.03 \times 10^{-2} L$

                          = $0.181 \times 10^{-2} mol$

From the balanced equation,  we get to know that  2 moles of $NO^{-}_{3}$ reacts with 3 moles of $Sn^{2+}$ .

$0.181 \times 10^{-2}$ moles of $NO^{-}_{3}$ reacts with M moles of $Sn^{2+}$.

        M = $\frac{0.181 \times 10^{-2}mol}{2}$

            = $0.09 \times 10^{-2} mol$

It is known that molar mass of tin is 118.71 g/mol. Hence, mass of Sn reacted will be as follows.

                 m = $0.09 \times 10^{-2}mol \times 118.71 g/mol$

                     = $10.68 \times 10^{-2} g$

So, percent mass of tin in the original sample = $\frac{\text{mass of tin reacted}}{\text{mass of sample}} \times 100$

                         = $\frac{10.68 \times 10^{-2} g}{0.528 g} \times 100$  

                         = 20.23 %

Thus, we can conclude that mass of tin is 20.23 %.