Question

A sample of impure tin of mass 0.526 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 3.67×10−2 L of the NO3− solution.
Find the percent by mass of tin in the original sample, assuming that it contains no other reducing agents.

Answer 1

Answer:

55.7%

Explanation:

The reaction that takes place is:

• 3Sn²⁺ + 2NO₃⁻ + 8H⁺ → 2NO + 3Sn⁺⁴ + 4H₂O

With the volume and concentration of NO₃⁻ solution, we can calculate the moles of Sn²⁺ that reacted:

• 3.67x10⁻² L * 0.0448 M = 1.64x10⁻³ mol NO₃⁻

• 1.64x10⁻³ mol NO₃⁻ * $\frac{3molSn^{2+}}{2molNO_{3}^{-}}$ = 2.47x10⁻³mol Sn²⁺

Now we convert moles of Sn to mass, using its atomic weight:

• 2.47x10⁻³mol Sn²⁺ * 118.71 g/mol = 0.293 g Sn

Finally we calculate the percent by mass of Sn:

• 0.293 g / 0.526 g * 100% = 55.7%