Question

$0.02\mathrm{M}$ ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ when reacts with $0.0288\mathrm{g}$ Ferrous Oxalate in an acidic medium. What is its volume(in ml)? $(\mathrm{Molar}\mathrm{mass}\mathrm{of}\mathrm{Fe}=56\mathrm{g}{\mathrm{mol}}^{-1})$

Answer 1

Step 1: Write the chemical equation

$$\begin{gathered} {\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7\left(\mathrm{s}\right)+2{\mathrm{FeC}}_2{\mathrm{O}}_4\left(\mathrm{s}\right)+7{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{l}\right)\to {\mathrm{Cr}}_2{\left({\mathrm{SO}}_4\right)}_3\left(\mathrm{s}\right)+{\mathrm{Fe}}_2{\left({\mathrm{SO}}_4\right)}_3\left(\mathrm{s}\right)+{\mathrm{K}}_2{\mathrm{SO}}_4\left(\mathrm{s}\right)+4{\mathrm{CO}}_2\left(\mathrm{g}\right)+7{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right) \\ \mathrm{Potassium}\mathrm{Ferrous}\mathrm{Sulphuric}\mathrm{Chromium}\mathrm{Ferric}\mathrm{Potassium}\mathrm{Carbon}\mathrm{Water} \\ \mathrm{Dichromate}\mathrm{Oxalate}\mathrm{Acid}\mathrm{Sulphate}\mathrm{Sulphate}\mathrm{Sulphate}\mathrm{Dioxide} \end{gathered}$$

Step 2: Milliequivalent of ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ is equal to the Milliequivalent of ${\mathrm{FeC}}_2{\mathrm{O}}_4$

${\mathrm{N}}_1{\mathrm{V}}_1={\mathrm{N}}_2{\mathrm{V}}_2$

where N is normality and V is volume

${\mathrm{n}}_1{\mathrm{M}}_1{\mathrm{V}}_1={\mathrm{n}}_2{\mathrm{M}}_2{\mathrm{V}}_2$

where n is the number of moles of electrons exchanged in pure reaction

Number of electrons exchanged for $\mathrm{Cr}$ in $\begin{array}{rcl}{\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7& =& 2+2x-2\times 7\\ & =& 2+2x-14\\ & =& 2x-12(x=6)\end{array}$

Number of electrons exchanged for $\mathrm{Cr}$ in $\begin{array}{rcl}{\mathrm{Cr}}_2{\left({\mathrm{SO}}_4\right)}_3& =& 2x-6=0\\ & =& 2x=6\\ & =& x\\ & =& 3\end{array}$

So the oxidation number from ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7\left(+6\right)\mathrm{to}{\mathrm{Cr}}_2{\left({\mathrm{SO}}_4\right)}_3\left(+3\right)$

The n factor change is $6$

Step 3: Number of electrons exchanged for $\mathrm{Fe}\mathrm{in}$ ${\mathrm{FeC}}_2{\mathrm{O}}_4$

$$\begin{gathered} x-2=0 \\ x=2 \end{gathered}$$

Number of electrons exchanged for $\mathrm{Fe}$ in ${\mathrm{Fe}}_2{\left({\mathrm{SO}}_4\right)}_3$$$\begin{gathered} 2x-6=0 \\ x=3 \end{gathered}$$

Number of electrons exchanged for $\mathrm{C}$in ${\mathrm{FeC}}_2{\mathrm{O}}_4$=$$\begin{gathered} \mathrm{2+2}\mathrm{x}-8=0 \\ 2\mathrm{x}-6=0 \\ \mathrm{x}=3 \end{gathered}$$

Number of electrons exchanged for $\begin{array}{rcl}\mathrm{C}\mathrm{in}{\mathrm{CO}}_2& =& \mathrm{x}-4=0\\ & =& \mathrm{x}=4\end{array}$

So n factor$=3$

Step 4: use formula

${\mathrm{n}}_1{\mathrm{m}}_1{\mathrm{V}}_1={\mathrm{n}}_2{\mathrm{m}}_2{\mathrm{V}}_2$

Molecular weight of ${\mathrm{FeC}}_2{\mathrm{O}}_4=144$

putting values in the above equation:

$\begin{array}{rcl}6\times 0.02\times {\mathrm{V}}_1& =& 3\times \frac{0.288}{144\times 1}\times 1000\\ V_1& =& 3\times \frac{0.0288}{144}\times \frac{1000\times 100}{6\times 0.02\times 1000}\\ {\mathrm{V}}_1& =& 50\mathrm{ml}\end{array}$

Therefore, $50\mathrm{ml}$ is the volume of ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$(in ml).