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"#