Given that, the sample of a substance who only oxidizable material is tin in the +2 state and we are meant to the determine the percentage composition of tin in the substance;
Then, the percentage composition of the tin in the substance is calculated as 172.84%
From the given information:
The molecular weight is the addition of all the atomic masses of a compound.
The molecular weight of K₂Cr₂O₇ is calculated as follows:
= (2 × 39.1) + (2 × 52) + (7 × 16)
= 294.2
The molecular weight of Sn²⁺ = 118.7
The oxidation half-reaction for Tin(ii) ion is as follows:
[tex]\mathbf{Sn^{2+}\to Sn^{4+}+2e^{-}}[/tex]
From the above equation, Tin (ii) ion oxidizes to Tin (iV) ion
The reduction of K₂Cr₂O₇ In the acidic medium, reduce Cr₂ to Chromium Cr³⁺.
[tex]\mathbf{K_2 Cr_2O_7 + 14H^{+}+6e^- \to 2K^+ + 2r^{3+} + 7H_2O}[/tex]
To calculate the percentage % of Tin in the sample, we must first know the Equivalent weight of each substance.
Calculating the equivalent weight (Eq. weight) of each substance in the samples, we have the following:
For K₂Cr₂O₇:
[tex]\mathbf{Eq. weight = \dfrac{molecular \ weight }{numbers \ of \ electrons \ lost \ or \ gained }}[/tex]
[tex]\mathbf{Eq. weight \ of \ K_2Cr_2O_7 = \dfrac{294.2 }{6}}[/tex]
[tex]\mathbf{Eq. weight \ of \ K_2Cr_2O_7 = 49.03 \ g/eq}[/tex]
[tex]\mathbf{Eq. weight \ of \ Sn^{2+} = \dfrac{118.7 }{2}}[/tex]
[tex]\mathbf{Eq. weight \ of \ Sn^{2+} =59.35 g/eq}[/tex]
The next step is to determine the Normality of the Solution.
Normality refers to the mole number per liter of the solution
[tex]\mathbf{Normality \ of \ K_2Cr_2O_7 = \dfrac{1.226 \ g }{49.03}\times \dfrac{1000}{250}}[/tex]
[tex]\mathbf{= \dfrac{4.904}{49.03} N}[/tex]
So, 23.90 mL of [tex]\mathbf{ \dfrac{4.904}{49.03} N}[/tex] of K₂Cr₂O₇ is required to be present in the solution.
= [tex]\mathbf{=23.90 mL \times \dfrac{4.904}{49.03 } \ of \ Sn \ ion}[/tex]
Thus, the amount of Sn in the sample is calculated as:
[tex]= \mathbf{\dfrac{4.904}{49.03}\times \dfrac{59.35}{1000}\times 23.90 \ mL}[/tex]
= 0.1419 g
Finally, the percentage composition of tIn (Sn) in the sample is:
[tex]\mathbf{= \dfrac{0.1419}{0.0821}\times 100\%}[/tex]
= 172.84%
NOTE: A 0.0821g sample of the substance was given. Ideally, it should be 0.821g. But the calculation was based on the data given.
Therefore, from the above explanation, we can conclude that the percentage % of Sn in the sample is 172.84%
Learn more about Redox reactions here:
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