Respuesta :
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
[tex] 6Fe^{2+} (aq) + Cr_{2} O_{7} ^{2-} (aq)+14H^{+} (aq)--> 6Fe^{3+} (aq)+2Cr^{3+} (aq) +7H2O(l) [/tex]
Let us use the information from part a to find grams of Fe.
Step 1 : Find moles of dichromate solution
[tex] 18 mL*\frac{1L}{1000mL} *\frac{0.225mol}{L}= 0.00405mol (dichromate) [/tex]
Step 2 : Find moles of Fe(II) solution using balanced equation and mol dichromate
[tex] 0.00405mol(Cr_{2}O_{7}^{2-}})*\frac{6molFe^{2+}}{1molCr_{2}O_{7}^{2-}} = 0.0243molFe^{2+} [/tex]
Step 3 : Convert moles of Fe(II) to grams of Fe
[tex] 0.0243molFe^{2+} *\frac{55.85g}{1mol} =1.357gFe [/tex]
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
[tex] percent (Fe) = \frac{Grams(Fe)}{Grams (Ore)} *100 [/tex]
% Fe by mass = [tex] \frac{1.357g}{4.05g} *100 [/tex]
% Fe by mass = 33.5%
Percentage of Fe by mass in the ore is 33.5%
