Answer :
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:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] .....(1)
- For NaOH:
Given mass of NaOH = 60 g
Molar mass of NaOH = 40 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of NaOH}=\frac{60g}{40g/mol}=1.5mol[/tex]
The chemical equation for the reaction of iron (III) chloride and NaOH follows:
[tex]FeCl_3+3NaOH\rightarrow Fe(OH)_3+3NaCl[/tex]
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 = [tex]\frac{1}{3}\times 1.5=0.5mol[/tex] 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:
[tex]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[/tex]
Hence, the mass of iron (III) chloride that must be added is 81.1 grams