The iron in a 1.805 -g sample containing some Fe2O3 is reduced to Fe2+ . The Fe2+ is titrated with 23.46 mL of 0.0195 M KMnO4 in an acidic solution (potassium is a spectator ion in this reaction). Calculate the percentage of Fe in the sample.
The unbalanced equation for the reaction is
Fe2+ + MnO4- 鈫?Fe3+ + Mn2+
________%Calculate percentage? ?
5 Fe2+ + MnO4- + 8H+ %26gt;%26gt; 5 Fe3+ + Mn2+ + 4 H2O
Moles KMnO4 = 0.02346 L x 0.0195 M = 0.000457
Moles Fe2+ = 5 x 0.000457 = 0.00229
= moles Fe3+
Moles Fe2O3 = 0.00229 / 2 = 0.00114
Mass Fe2O3 = 0.00114 x 159.694 g/mol= 0.1821
% = 0.01821 x 100 / 1.805 =10.09
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