Respuesta :
Solution :
[tex]$Fe(OH)_3+3CH_3COOK \rightarrow Fe(CH_3COO)_3 + 3KOH$[/tex]
(iron (potassium
hydroxide) acetate)
Number of moles of [tex]$Fe(OH)_3 = \frac{\text{mass of }Fe(OH)_3}{\text{molar mass of }Fe(OH)_3}$[/tex]
[tex]$=\frac{2.143 \ g}{106.867 \ g/mol}$[/tex]
= 0.02005 mol
Number of moles of [tex]$CH_3COOK = \frac{\text{mass of }CH_3COOK}{\text{molar mass of }CH_3COOK}$[/tex]
[tex]$=\frac{5.478 \ g}{98.15 \ g/mol}$[/tex]
= 0.0585 mol
Since 1 mol of [tex]$Fe(OH)_3$[/tex] reacts with 3 mols of [tex]$CH_3COOK$[/tex]
Therefore, number of moles of [tex]$CH_3COOK$[/tex] reacted [tex]$=\frac{0.0585}{3 }$[/tex] mol
= 0.0195 mol
Therefore the limiting reagent is [tex]$CH_3COOK$[/tex] and hence the number of moles of [tex]$Fe(CH_3COO)_3$[/tex] produced is 0.0195 mol
Amount of [tex]$Fe(CH_3COO)_3$[/tex] produced = moles x molar mass
= 0.0195 moles x 232.98 g/mol
= 4.5431 g
