# Question #e9551

Feb 1, 2016

$32.40 \text{ml}$

#### Explanation:

$C {r}_{2} {O}_{7}^{2 -} + 14 {H}^{+} + 6 e \rightarrow 2 C {r}^{3 +} + 7 {H}_{2} O \text{ } \textcolor{red}{\left(1\right)}$

$C u \rightarrow C {u}^{2 +} + 2 e \text{ } \textcolor{red}{\left(2\right)}$

So to get the electrons to balance we multiply $\textcolor{red}{\left(2\right)}$ by 3 then add both sides $\Rightarrow$

$C {r}_{2} {O}_{7}^{2 -} + 14 {H}^{+} + \cancel{6 e} + 3 C u \rightarrow 2 C {r}^{3 +} + 7 {H}_{2} O + 3 C {u}^{2 +} + \cancel{6 e}$

From this you can see that 1 mole of $C {r}_{2} {O}_{7}^{2 -}$ will oxidise 3 moles of $C u$.

We are given the mass of copper so we can find the number of moles by dividing by the mass of 1 mole:

${n}_{C u} = \frac{m}{M} _ r = \frac{5.25}{63.54} = 0.0826$

$\therefore n C {r}_{2} {O}_{7}^{2 -} = \frac{0.0826}{3} = 0.02754$

We know that $c = \frac{n}{v}$

$\therefore v = \frac{n}{c} = \frac{0.02754}{0.85} = 0.0324 \text{L}$

or $32.40 \text{ml}$