Question

Question #e9551

Answer 1

`32.40"ml"`

Explanation:

Start with the 1/2 equations:

`Cr_2O_7^(2-)+14H^(+)+6erarr2Cr^(3+)+7H_2O" "color(red)((1))`

`CurarrCu^(2+)+2e" "color(red)((2))`

So to get the electrons to balance we multiply `color(red)((2))` by 3 then add both sides `rArr`

`Cr_2O_7^(2-)+14H^(+)+cancel(6e)+3Curarr2Cr^(3+)+7H_2O+3Cu^(2+) + cancel(6e)`

From this you can see that 1 mole of `Cr_2O_7^(2-)` will oxidise 3 moles of `Cu`.

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

`n_(Cu)=m/M_r=5.25/63.54=0.0826`

`:.nCr_2O_7^(2-)=0.0826/3=0.02754`

We know that `c=n/v`

`:. v=n/c=0.02754/0.85=0.0324"L"`

or `32.40"ml"`