Question

The fe2+(aq) solution described in part a was produced by dissolving 4.05 g of ore in acid. what was the percentage of fe by mass in this ore sample? express your answer to three significant figures and include the appropriate units. submithintsmy a

Answer 1

Information from part a is missing in the question which is given below.

Part a ) The titration of 25.0 mL of an iron(II) solution required 18.0 mL of a 0.225 M solution of dichromate to reach the equivalence point. What is the molarity of the iron(II) solution?

Balanced chemical equation for the reaction is

`6Fe^(2+) (aq) + Cr_(2) O_(7) ^(2-) (aq)+14H^(+) (aq)--> 6Fe^(3+) (aq)+2Cr^(3+) (aq) +7H2O(l)`

Let us use the information from part a to find grams of Fe.

Step 1 : Find moles of dichromate solution

`18 mL*(1L)/(1000mL) *(0.225mol)/(L)= 0.00405mol (dichromate)`

Step 2 : Find moles of Fe(II) solution using balanced equation and mol dichromate

`0.00405mol(Cr_(2)O_(7)^(2-)})*(6molFe^(2+))/(1molCr_(2)O_(7)^(2-)) = 0.0243molFe^(2+)`

Step 3 : Convert moles of Fe(II) to grams of Fe

`0.0243molFe^(2+) *(55.85g)/(1mol) =1.357gFe`

Step 4 : Find % of Fe by mass

Grams of Fe present in the ore are 1.357

Grams of ore are given as 4.05

% of Fe in the ore can be found out using following formula

`percent (Fe) = (Grams(Fe))/(Grams (Ore)) *100`

% Fe by mass =
`(1.357g)/(4.05g) *100`

% Fe by mass = 33.5%

Percentage of Fe by mass in the ore is 33.5%