Iron (III) ion reacts with sulfide. An iron solution is made by dissolving 1.25 g of iron sulfate in water to produce 345.0 mL of solution

Question

cont. 12.0 mL of this solution was mixed with 18.0 mL of 0.113 M sodium sulfide,

and a precipitate of iron sulfide formed. Whats the concentration of the Iron sulfate solution

Answer 1

Okay, let's first calculate the number of moles of FeSO4 dissolved...

1.25 g FeSO4/151.917 g/mol = 0.008228 mol FeSO4

Then, we will calculate the molarity in moles/L

0.008228 mol FeSO4/0.345 L = 0.02385 M (or moles/L). This is also the same concentration of Fe3+ and SO42- (Iron 3 ion and sulfate ion)

If only 12.0 mL of this solution is used, we can then calculate the number of moles of Fe3+ using M x V = moles: 0.02385 mol/L x 0.012 L = 0.000286 moles Fe3+.

A table of solubility rules will state that iron and sulfide will react by forming a precipitate. Let's see how many moles of sulfide are added to the iron solution by using the numbers for volume and molarity given:

0.113 moles/L x 0.018 L = 0.002034 moles sulfide (or S2- ion).

Iron 3+ and sulfide will react in a 2:3 ratio to form the precipitate:

2 Fe3+ (aq) + 3 S2- (aq)- −−−> Fe2S3 (s)

Ha ha! They tried to trick me, The moles of iron and sulfide are present in a 2:3 ratio, but not in the fashion required. Ideally, there would be 1.5 times as many moles of sulfide as iron (as we can tell by looking at the ratio of the coefficients in the balanced net ionic equation). But! In this case, there actually 1.5 times as many moles or iron as sulfide. This will make sulfide a limiting reactant so all of the moles of sulfide will be consumed. The total volume of the added solutions will be 12.0 mL + 18.0 mL = 30.0 mL.

Finally, let's calculate the number of moles Fe2S3 from the balanced net ionic equation starting with the limiting moles of sulfide.

0.002034 moles S2- x (1 mol Fe2S3/3 mol S2-) = 0.000678 moles Fe2S3.

To answer this seemingly never-ending question, we calculate the concentration of Fe2S3 in moles per liter.

0.000678 moles/0.030 L = 0.0226 moles per L (or 0.00226 M)