Asample 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.

55.7%

Explanation:

The reaction that takes place is:

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

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

Finally we calculate the percent by mass of Sn:

if the concentration of products is decreased, the equilibrium will shift toward the products). add an inert gas (one that is not involved in the reaction) to the constant-volume reaction mixture: this will increase the total pressure of the system, but will have no effect on the equilibrium condition.

explanation: