Excess of NaOH, (aq) was added to 100, mL of | vedclass

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(A) The balanced chemical equation is: $FeCl_3 (aq) + 3NaOH (aq) 	o Fe(OH)_3 (s) + 3NaCl (aq)$.Molar mass of $Fe(OH)_3 = 56 + 3 	imes (16 + 1) = 56

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$0.2$

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(A) The balanced chemical equation is: $FeCl_3 (aq) + 3NaOH (aq) 	o Fe(OH)_3 (s) + 3NaCl (aq)$.Molar mass of $Fe(OH)_3 = 56 + 3 	imes (16 + 1) = 56 + 51 = 107\, g\, mol^{-1}$.Moles of $Fe(OH)_3 = \frac{	ext{mass}}{	ext{molar mass}} = \frac{2.14\, g}{107\, g\, mol^{-1}} = 0.02\, mol$.From the stoichiometry of the reaction,$1\, mole$ of $FeCl_3$ produces $1\, mole$ of $Fe(OH)_3$.Therefore,moles of $FeCl_3$ present = $0.02\, mol$.Volume of $FeCl_3$ solution = $100\, mL = 0.1\, L$.Molarity of $FeCl_3 = \frac{	ext{moles of solute}}{	ext{volume of solution in } L} = \frac{0.02\, mol}{0.1\, L} = 0.2\, M$.

Excess of $NaOH\, (aq)$ was added to $100\, mL$ of $FeCl_3\, (aq)$ resulting in $2.14\, g$ of $Fe(OH)_3$. The molarity of $FeCl_3\, (aq)$ is (Given molar mass of $Fe=56\, g\, mol^{-1}$ and molar mass of $Cl=35.5\, g\, mol^{-1}$) (in $, M$)

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