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

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

Step-by-step 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 = (Mass\ taken)/(Molar\ mass)`

Thus,

`Moles= (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+))* Volume_(Sn^(2+))`

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

So,

`Molarity_(Sn^(2+))=(2* 0.00424)/(0.0244)`

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