Respuesta :
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 [tex]Sn^{2+}[/tex]. The solution is then titrated with a 0.0448 M solution of [tex]NO_3^-[/tex], which is reduced to NO(g). The equivalence point is reached upon the addition of [tex]4.13\times 10^{-2 }L[/tex] of the [tex]NO_3^-[/tex] 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 [tex]NO_3^-[/tex].
[tex]\text{Moles of }NO_3^-=\text{Concentration of }NO_3^-\times \text{Volume of solution}[/tex]
[tex]\text{Moles of }NO_3^-=0.0448M\times 4.13\times 10^{-2}L=1.85\times 10^{-3}mol[/tex]
Now we have to calculate the moles of [tex]Sn^{4+}[/tex]
The balanced chemical reaction is,
[tex]2NO_3^-(aq)+3Sn^{2+}(aq)+8H^+(aq)\rightarrow 2NO(g)+3Sn^{4+}(aq)+4H_2O(l)[/tex]
From the reaction, we conclude that
As, 2 mole of [tex]NO_3^-[/tex] react with 3 mole of [tex]Sn^{2+}[/tex]
So, [tex]1.85\times 10^{-3}[/tex] moles of [tex]NO_3^-[/tex] react with [tex]\frac{1.85\times 10^{-3}}{2}\times 3=2.78\times 10^{-3}[/tex] moles of [tex]Sn^{2+}[/tex]
Now we have to calculate the mass of [tex]Sn^{2+}[/tex]
[tex]\text{ Mass of }Sn^{2+}=\text{ Moles of }Sn^{2+}\times \text{ Molar mass of }Sn^{2+}[/tex]
Molar mass of tin = 118.71 g/mol
[tex]\text{ Mass of }Sn^{2+}=(2.78\times 10^{-3}moles)\times (118.71g/mole)=0.330g[/tex]
Mass of [tex]Sn^{2+}[/tex] reacted = 0.330 g
Original mass of [tex]Sn^{2+}[/tex] = 0.538 g
Now we have to calculate the percent mass of tin in the original sample.
[tex]\% \text{ mass of tin in the original sample}=\frac{\text{ Reacted mass of }Sn^{2+}}{\text{ Original mass of }Sn^{2+}}\times 100[/tex]
[tex]\% \text{ mass of tin in the original sample}=\frac{0.330g}{0.538g}\times 100=61.3\%[/tex]
Therefore, the percent mass of tin in the original sample is 61.3 %
