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?

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:

    .....(1)

Given mass of NaOH = 60 g

Molar mass of NaOH = 40 g/mol

Putting values in equation 1, we get:

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

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 = 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:

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

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