Question

1. Can 0.250 mol NaOH consume all of the available FeCl3 (0.308 mol)? FeCl3(aq) + 3 NaOH(aq) → Fe(OH)3(s) + 3 NaCl(aq)

2. Now consider the reaction when 45.0 g NaOH have been added. What amount of NaOH is this, and what amount of FeCl₃can be consumed by it?

Answer 1

1.

The given balanced reaction between FeCl3 and NaOH is

Note that 1 mol of FeCl3 completely reacts with 3 moles of NaOH.

Hence, the number of moles of NaOH that will react with the available 0.308 mol of FeCl3 is

Hence, 0.924 mol of NaOH is needed to consume all the FeCl3 available.

It means, 0.250 mol of NaOH, which is less than the amount required, will not be able to consume all the available FeCl3.

2.

Amount of NaOH added = 45.0 g

Molar mass of NaOH = 39.997 g/mol

Hence, number of moles of NaOH in 45.0 g is

We already know from the balanced reaction that 3 moles of NaOH can consume 1 mol of FeCl3.

Hence, the number of moles of FeCl3 that can be consumed by 1.125 mol of NaOH is

Molar mass of FeCl3 = 162.2 g/mol

Hence, mass of 0.375 mol of FeCl3 is

Hence, 45.0 g of NaOH can consume 0.375 mol of FeCl3 (approximately 60.8 g of FeCl3)

It means if we add 45.0 g of NaOH to the available 0.308 mol of FeCl3, all the FeCl3 will be consumed with some NaOH leftover.