Based on the balanced chemical equation shown below, determine the mass percent of fe3+ in a 0.strings_1450g 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) → 2 fe2+(aq) + sn4+(aq).

I think there is a mistake in strings mass, I think it is 0.7450 g, not 0.1450 g
First, we need to get no.of moles of SN2+ = molarity * volume

= 0.1 M* 0.0224 L = 0.00224 moles 

So, according to the reaction equation:

2Fe3+(aq)  +  SN2+(aq) → 2 Fe2(aq) + SN4+ (Aq)

we can see that the molar ratio between Fe3+ and SN2+ = 2:1

∴ to get the no.of moles Fe3+ we have to multiply SN2+ by 2
= 2 * 0.00224 = 0.00448 moles 

now, when we know the molar mass of  Fe3+ = 55.847 g/mol

so, we can use it and the no. of moles to get the mass

mass Fe3+ =  moles * molar mass
                   = 0.00448 * 55.847 = 0.25 g

∴ Mass Fe3+ % = 0.25/0.7450 * 100 = 33.6 %

The answer options is d