Question

A wire weighing 0.250 g and containing 94.75% Fe is dissolved in HCl. The iron is completely oxidized to Fe3+ by bromine water. The solution is then treated with tin(II) chloride to bring about the reaction Sn2+(aq) + 2Fe3+(aq) → 2Fe2+(aq) + Sn4+(aq) + H2O(l) If 24.4 mL of tin(II) chloride solution is required for complete reaction, what is the molarity of the tin(II) chloride solution

Answer 1

Answer:

Molarity of the tin(II) chloride solution = 0.3475 M

Explanation:

Iron is present in the wire by a percentage of 94.75% .

Thus,

Percent composition is percentage by the mass of element present in the compound.

Given that the $Fe$ is 94.75 % by mass.

100 g of the wire contains 94.75 g of $Fe$

0.250 g of the wire contains 0.9475*0.250 g of $Fe$

Mass of Fe = 0.2369 g

Molar mass of Fe = 55.845 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{0.2369\ g}{55.845\ g/mol}$

Moles of iron = 0.00424 moles

According to the reaction,  

$Sn^{2+}_{(aq)}+2Fe^{3+}_{(aq)}\rightarrow 2Fe^{2+}_{(aq)}+Sn^{4+}_{(aq)} + H_2O_{(l)}$

At equivalence point

Moles of $Sn^{2+}$ = 2*Moles of $Fe^{3+}$

Considering :-

Moles of $Sn^{2+}=Molarity_{Sn^{2+}}\times Volume_{Sn^{2+}}$

Volume = 24.4 mL = 0.0244 L ( 1 mL = 0.001 L)

So,

$Molarity_{Sn^{2+}}=\frac{2\times 0.00424}{0.0244}$

Molarity of the tin(II) chloride solution = 0.3475 M