Question #e9551

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Answer 1 · 2016-02-01T18:11:37

$32.40"ml"$

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 $32.40"ml"$