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 %