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Schach [20]
3 years ago
14

A different brand of iron tablet was analysed by Hitration with 0.0093 mol.L" potassium

Chemistry
1 answer:
Gre4nikov [31]3 years ago
4 0

The concentration of iron used in the titration : 0.009 M

<h3>Further explanation</h3>

Given

Reaction

Cr₂O₇²⁻ + 6Fe²⁺ + 14H⁺ ⇒ 2Cr³⁺ + 6Fe³⁺ + 7H₂O

0.0093 mol/L potassium  dichromate

200 cm³ of dilute acid, 25cm³   was used in the titration.

Required

the concentration of iron

Solution

Titration formula

C₁V₁n₁=C₂V₂n₂⇒ From equation : n₁=6n₂(1=Cr₂O₇, 2=Fe)

titration average : 33+32.05+32.15+32.1 / 4 = 32.325 cm³(ml)

25 cm³ of iron solution used in titration :

\tt V_1=32.325~ml\\\\V_2=25~ml\\\\C_1=0.0093~M\\\\\\\\C_1.V_1.n_2=C_2.V_2.n_1\\\\0.0093\times 32.325\times 6=C_2\times 25\times 1\\\\C_2=0.07215~M

Dilution(25 ml from 200 ml iron solution)

\tt C_2.V_2=C_1.V_1\\\\0.007215\times 25=C_1\times 200\\\\C_1=0.009~M

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There is some info missing. I think this is the original question.

<em>A chemist adds 370.0mL of a 2.25 M iron(III) bromide (FeBr₃) solution to a reaction flask. Calculate the mass in kilograms of iron(III) bromide the chemist has added to the flask. Be sure your answer has the correct number of significant digits.</em>

<em />

We have 370.0 mL of 2.25 M iron(III) bromide (FeBr₃) solution. The moles of FeBr₃ are:

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