For Reaction 7-1,how many milliliters of$\mathbf{0}\mathbf{.}\mathbf{1650}\mathbf{M}\mathbf{}{\mathbf{KMnO}}_{\mathbf{4}}$are needed to react with $\mathbf{108}\mathbf{.}\mathbf{0}\mathbf{mL}$of $\mathbf{0}\mathbf{.}\mathbf{1650}\mathbf{M}$oxalic acid? How many milliliters of $\mathbf{0}\mathbf{.}\mathbf{1650}\mathbf{M}$oxalic acid are required to react with $\mathbf{108}\mathbf{.}\mathbf{0}\mathbf{mL}\mathbf{}\mathbf{of}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{1650}{\mathbf{MKMnO}}_{\mathbf{4}}$ ?

Reaction 7-1 is expressed below:

$5{\mathrm{C}}_2{\mathrm{O}}_4{\mathrm{H}}_2+2{{\mathrm{MnO}}_4}^{-}+6{\mathrm{H}}^{+}\to 10{\mathrm{CO}}_2+2{\mathrm{Mn}}^{2+}+8{\mathrm{H}}_2\mathrm{O}$

The mole ratio is the ratio of the moles of any two chemical entities in a compound or chemical process.

$$\begin{gathered} \mathrm{Concentration}\mathrm{of}{\mathrm{KMnO}}_4=0.165\mathrm{M} \\ \mathrm{Concentration}\mathrm{of}{\mathrm{C}}_2{\mathrm{O}}_4{\mathrm{H}}_2=0.165\mathrm{M} \\ \mathrm{Volume}\mathrm{of}{\mathrm{C}}_2{\mathrm{O}}_4{\mathrm{H}}_2=108\mathrm{mL}=0.108\mathrm{L} \\ \mathrm{First},\mathrm{let}\text{'}\mathrm{s}\mathrm{find}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{moles}\mathrm{of}{\mathrm{C}}_2{\mathrm{O}}_4{\mathrm{H}}_2: \\ \mathrm{Moles}=\mathrm{Concentration}\times \mathrm{Volume} \\ =0.165\mathrm{M}\times 0.108\mathrm{L} \\ =0.01782\mathrm{mol} \end{gathered}$$

And since the mole ratio between ${\mathrm{C}}_2{\mathrm{O}}_4{\mathrm{H}}_2$and ${{\mathrm{MnO}}_4}^{-}$is $5:2$, the number of moles of ${{\mathrm{MnO}}_4}^{-}$

$0.01782{\mathrm{molC}}_2{\mathrm{O}}_4{\mathrm{H}}_2\times \frac{2{\mathrm{molMnO}}_{-}^{-}4}{5{\mathrm{molC}}_{-}{\mathrm{O}}_{-}4{\mathrm{H}}_{-}2}=0.007128{\mathrm{molMnO}}_{-}^{-}4$

Because the mole ratio of ${\mathrm{MnO}}_4^{-}$and $\mathrm{KMnO}$is $1:1_2{\mathrm{KMnO}}_4$will have a mole number of $0.007128\mathrm{mo}\mathrm{l}.$

So, given the number of moles and the concentration of ${\mathrm{KMnO}}_4$you can calculate the volumes required:

$$\begin{gathered} \mathrm{Volume}\mathrm{of}{\mathrm{KMnO}}_4=\frac{\mathrm{Moles}}{\mathrm{concentration}} \\ =\frac{0.007128\mathrm{mol}}{0.165\mathrm{M}} \\ =0.0432\mathrm{L} \\ =43.2\mathrm{mL} \\ \mathrm{Thus},\mathrm{volume}\mathrm{of}{\mathrm{KMnO}}_4\mathrm{needed}\mathrm{is}43.2\mathrm{mL}. \end{gathered}$$

$$\begin{gathered} \mathrm{Concentration}\mathrm{of}{\mathrm{KMnO}}_4=0.165\mathrm{M} \\ \mathrm{Concentration}\mathrm{of}{\mathrm{C}}_2{\mathrm{O}}_4{\mathrm{H}}_2=0.165\mathrm{M} \\ \mathrm{To}\mathrm{begin},\mathrm{determine}\mathrm{the}\mathrm{amount}\mathrm{of}\mathrm{moles}\mathrm{of}:{\mathrm{KMnO}}_4 \end{gathered}$$

$$\begin{gathered} \mathrm{Moles}=\mathrm{Concentration}\times \mathrm{Volume} \\ =0.165\mathrm{M}\times 0.108\mathrm{L} \\ =0.01782\mathrm{mol} \end{gathered}$$

And, because the mole ratio of $\mathrm{KMnO}4$and ${\mathrm{MnO}}_4^{-}$is $1:1$the number of moles of ${\text{MnO}}_4^{-}$ will also be $0.01782$

Because the mole ratio of ${\mathrm{MnO}}_4^{-}$and${\mathrm{C}}_2{\mathrm{O}}_4{\mathrm{H}}_2$is $2:5$,the number of moles of ${\mathrm{C}}_2{\mathrm{O}}_4{\mathrm{H}}_2$: is calculated as follows:

$0.01782{\mathrm{molMnO}}_4\times \frac{5{\mathrm{molC}}_2{\mathrm{O}}_4{\mathrm{H}}_2}{2{\mathrm{molMnO}}_4}=0.04455{\mathrm{molC}}_2{\mathrm{O}}_4{\mathrm{H}}_2$

With the number of moles and concentration of ${\mathrm{C}}_2{\mathrm{O}}_4{\mathrm{H}}_2$,find the volumes needed:

${\mathrm{C}}_2{\mathrm{O}}_4{\mathrm{H}}_2$=role="math" localid="1654850856154" $$\begin{gathered} =\frac{\mathrm{Moles}}{\mathrm{concentration}} \\ =\frac{0.04455\mathrm{mol}}{0.165\mathrm{M}} \\ =0.27\mathrm{L} \\ =270\mathrm{mL} \end{gathered}$$

$\mathrm{Thus},\mathrm{volume}\mathrm{of}{C}_2{O}_4{H}_2\mathrm{needed}\mathrm{is}2.70\mathrm{mL}.$