What is the mass percent of Fe3+ in the sample?

Question

What is the mass percent of Fe3+ in the sample if a 0.6450 gram sample of iron ore requires 22.40 mL of 0.1000 M Sn2+ to completely react with the Fe3+ content of the ore? The reaction is...

2Fe3+(aq) + Sn 2+(aq) ---> 2 Fe2+ (aq) + Sn4+ (aq)


a.) 19.40 %
b.) 25.02 %
c.) 38.79 %
d.) 77.59%

please show work

Answer 1

1) We can compute first how many moles of Sn2+ are there in 22.4 mL in soluction 0.1M:

n Sn2+ = (0.0224 L)(0.1 moles/L) = 0.00224 moles Sn2+

2) According to reaction:

2Fe3+(aq) + Sn 2+(aq) ---> 2 Fe2+ (aq) + Sn4+ (aq)

0.00224 moles of Sn2* will react with:

(0.00224)(2 moles Fe3+) = 0.00448 moles Fe3+

3) Let´s compute the amount in grames knowing the moles of Fe3+ above, just remember that molecular mass of Fe = 55.84 grames/mol hence,
grames Fe3+ = (0.00448 moles)(55.84 g/mol) = 0.2501 g Fe3+

4) To calculate the percentage of Iron in sample (remember that sample weighs 0.6450 grames):

% Fe3+ = (0.2501 g / 0.6450 g) x100% = 38.78%

5) So, option c) is the right answer.

That´s it!

Good luck!

Answer 2

the moles of [Sn2+] n = C * V = 0.1 * 0.02240 = 22.4 * 10^(-4)
the moles of Fe 3+ that reacted 22.4*10^(-4) * 2 / 1= 44.8*10^(-4)
the amount of Fe 3+ 56 * 44.8 * 10^(-4) = 0.25088 g
the mass percent of Fe3+ 0.25088/0.6450 * 100 = 38.89%