Question

Now examine the reaction between iron(III) chloride and sodium hydroxide. Use 60 g NaOH and add FeCl3. What mass in grams of FeCl3 must be added to just consume 60 g NaOH?

Answer 1

Answer: The mass of iron (III) chloride that must be added is 81.1 grams

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$     .....(1)

• For NaOH:

Given mass of NaOH = 60 g

Molar mass of NaOH = 40 g/mol

Putting values in equation 1, we get:

$\text{Moles of NaOH}=\frac{60g}{40g/mol}=1.5mol$

The chemical equation for the reaction of iron (III) chloride and NaOH follows:

$FeCl_3+3NaOH\rightarrow Fe(OH)_3+3NaCl$

By Stoichiometry of the reaction:

3 moles of NaOH reacts with 1 mole of iron (III) chloride

So, 1.5 moles of NaOH will react with = $\frac{1}{3}\times 1.5=0.5mol$ of iron (III) chloride

Now, calculating the mass of iron (III) chloride from equation 1, we get:

Molar mass of iron (III) chloride = 162.2 g/mol

Moles of iron (III) chloride = 0.5 moles

Putting values in equation 1, we get:

$0.5mol=\frac{\text{Mass of iron (III) chloride}}{162.2g/mol}\\\\\text{Mass of iron (III) chloride}=(0.5mol\times 162.2g/mol)=81.1g$

Hence, the mass of iron (III) chloride that must be added is 81.1 grams