To 50 Litres of 0.2 N , 5 Litres of 1 N and 15 Litres of 0.1 N solution are added. The resulting pre — Redox Reactions and Volumetric Analysis Chemistry Question
Question
To 50 Litres of 0.2 N $NaOH$, 5 Litres of 1 N $HCl$ and 15 Litres of 0.1 N $FeCl_3$ solution are added. The resulting precipitate of iron(III) hydroxide is ignited to form $Fe_2O_3$. What is the mass of $Fe_2O_3$ obtained (in grams)? (Molar mass of $Fe_2O_3$ = 160 g/mol)
💡 Solution & Explanation
Equivalents of $NaOH$ = $50 \times 0.2 = 10$ eq. Equivalents of $HCl$ = $5 \times 1 = 5$ eq. Equivalents of $FeCl_3$ = $15 \times 0.1 = 1.5$ eq. $NaOH$ first neutralizes $HCl$ (consuming 5 eq), leaving $10 - 5 = 5$ eq of $NaOH$. $NaOH$ then reacts with $FeCl_3$ to precipitate $Fe(OH)_3$. Since 5 > 1.5, $FeCl_3$ is the limiting reagent. 1.5 equivalents of $Fe(OH)_3$ are formed. Upon ignition, it forms $Fe_2O_3$. The equivalents of $Fe_2O_3$ produced = 1.5 eq. The n-factor of $Fe_2O_3$ is $2 \times 3 = 6$. Equivalent weight of $Fe_2O_3 = 160 / 6 = 80/3$. Mass = $1.5 \times (80/3) = 40$ g.