Question

The amount of iron in steel can be determined by titration with dichromate. First, the steel is dissolved in hydrochloric acid to oxidize all the iron to Fe 2 + . An indicator is added that will react with excess dichromate. Then, potassium dichromate is added to the solution until the end point is reached. Cr 2 O 2 − 7 + 6 Fe 2 + + 14 H + ⟶ 2 Cr 3 + + 6 Fe 3 + + 7 H 2 O For the analysis of 2.00 g of steel, the titration required 21.20 mL of 0.1800 M K 2 Cr 2 O 7 . What was the mass percentage of iron in this steel sample?

Answer 1

Answer: 64.1%

Step-by-step explanation:

mL---> L * M(Cr2O7)

(21.20/1000)*0.1800=0.003816

6 mols of iron react with 1 mol of Cr2O7

6* 0.003816=0.022896

mols of iron * molar mass of iron

0.022896*56= 1.282176

grams of iron/ grams of sample * 100

(1.282176/2.00)*100= 64.1088% =64.1%

Answer 2

% Fe= 63.9

Step-by-step explanation:

The amount of iron that was titrated can be calculated as follows:

mol K₂Cr₂O₇ = 0.02120 L x 0.180 M = 3.82 x 10⁻³

mol Fe titrated = 3.78 mol Cr₂O₇²⁻ x 10⁻³ x 6 mol Fe²⁺/mol Cr₂O₇²⁻

= 2.29 x 10⁻²

mass Fe = 2.29 x 10⁻² mol Fe²⁺ x 55.845 g/mol Fe²⁺ = 1.28 g Fe

% Fe = 1.28 g Fe / 2.00 g steel x 100 g steel = 63.9 %