Answer:
38.9%
Explanation:
below, determine the mass percent of Fe3+ in a 0.6450 g sample of iron ore, if 22.40 mL of a 0.1000 M stannous chloride, SnCl2(aq), solution is required to completely react with the Fe3+ present in the ore sample. The chemical equation for the reaction is
2 Fe3+ (aq) + Sn2+ (aq) →2Fe2+ (aq) + Snº+(aq).
O 19.40%
O 6.196%
O 38.79%
O 9.697%
2 Fe3+ (aq) + Sn2+ (aq) →2Fe2+ (aq) + Snº+(aq).ole of Sn2+
for every mole of Sn2+, tere are 2 moles of Fe3+
so there are 0.00224 X 2= 0.00448 moles Fe3+
iron atomic mass is 56
56 X 0.00448 = 0.251 gm Fe3+
so the mass % of the Fe3+ in the 0.645 gm iron ore sample is
(0.251/0.645) X 100 = 38.9%