Awire 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

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 is 94.75 % by mass.

100 g of the wire contains 94.75 g of

0.250 g of the wire contains 0.9475*0.250 g of

Mass of Fe = 0.2369 g

Molar mass of Fe = 55.845 g/mol

The formula for the calculation of moles is shown below:

Thus,

Moles of iron = 0.00424 moles

According to the reaction,  

At equivalence point

Moles of = 2*Moles of

Considering :-

Moles of

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

So,

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

120,630,890

step-by-step explanation:

110+10=120,620+10=630,880+10=890