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

$18 mL*\frac{1L}{1000mL} *\frac{0.225mol}{L}= 0.00405mol (dichromate)$

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

$0.00405mol(Cr_{2}O_{7}^{2-}})*\frac{6molFe^{2+}}{1molCr_{2}O_{7}^{2-}} = 0.0243molFe^{2+}$

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

$0.0243molFe^{2+} *\frac{55.85g}{1mol} =1.357gFe$

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

$percent (Fe) = \frac{Grams(Fe)}{Grams (Ore)} *100$

% Fe by mass = $\frac{1.357g}{4.05g} *100$

% Fe by mass = 33.5%

Percentage of Fe by mass in the ore is 33.5%