Question

5.478 grams of potassium acetate and 2.143 grams of iron(III) hydroxide are added to a beaker containing 100.0 mL of water and stirred vigorously. A solid settles to the bottom of the beaker. If the water is decanted and the solid is dried, what is the maximum mass of solid that should be recovered

Answer 1

`$Fe(OH)_3+3CH_3COOK \rightarrow Fe(CH_3COO)_3 + 3KOH$`

(iron (potassium

hydroxide) acetate)

Number of moles of
`$Fe(OH)_3 = \frac{\text{mass of }Fe(OH)_3}{\text{molar mass of }Fe(OH)_3}$`

`$=(2.143 \ g)/(106.867 \ g/mol)$`

= 0.02005 mol

Number of moles of
`$CH_3COOK = \frac{\text{mass of }CH_3COOK}{\text{molar mass of }CH_3COOK}$`

`$=(5.478 \ g)/(98.15 \ g/mol)$`

= 0.0585 mol

Since 1 mol of
`$Fe(OH)_3$` reacts with 3 mols of
`$CH_3COOK$`

Therefore, number of moles of
`$CH_3COOK$` reacted
`$=(0.0585)/(3 )$` mol

= 0.0195 mol

Therefore the limiting reagent is
`$CH_3COOK$` and hence the number of moles of
`$Fe(CH_3COO)_3$` produced is 0.0195 mol

Amount of
`$Fe(CH_3COO)_3$` produced = moles x molar mass

= 0.0195 moles x 232.98 g/mol

= 4.5431 g