Question

A chemist prepares a precipitation titration using 10.00 mL of a 0.100M solution labeled as "iron chloride." This aliquot of the iron chloride sample is titrated using a a 0.0740M sodium hydroxide solution. The end point of the titration occurs when 27.03 mL of the 0.0740MNaOH solution is dispensed. Determine the ratio of the number of moles of sodium hydroxide to moles of iron chloride.

moles sodium hydroxide:_____mol

Answer 1

The ratio of the number of moles of sodium hydroxide to moles of iron chloride in the titration is 3:1, assuming the simplest form of iron chloride (FeCl3) is present.

Step-by-step explanation:

The ratio of the number of moles of sodium hydroxide to moles of iron chloride is 3:1.

First, we calculate the moles of sodium hydroxide (NaOH) that were used in the titration:

Moles NaOH = Volume (L) × Molarity (M)
= 0.02703 L × 0.0740 M
= 0.00200122 mol

Next, we calculate the moles of iron chloride (FeCln) in the 10.00 mL aliquot:

Moles FeCln = Volume (L) × Molarity (M)
= 0.01000 L × 0.100 M
= 0.00100 mol

To find the ratio, we assume the simplest formula, FeCl3, which reacts with NaOH in 1:3 stoichiometry:

Moles NaOH: Moles FeCl3 = 0.00200122 mol: 0.00100 mol = 3:1