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?

% Fe=  63.9

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 %

"standard molar enthalpy:

h2o ( liquid water ) : - 285.8 kj/mole

h2o ( water vapor ) : - 241.8 kj / mole.

δhrxn = - 241. 8 - ( - 285.8 )= -241.8 + 285.8 = 44.0

answer: 44.0 kj"

wher is the rest of the question