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Starting with 250 ml of Au^(3+) solution...

Starting with `250 ml` of `Au^(3+)` solution `250 ml` of `Fe^(2+)` solution, the following equilibrium is established `Au^(3+) (aq) +3Fe^(2+)(aq) hArr 3Fe^(3+) (aq) +Au(s)`
At equilibrium the equivalents of `Au^(3+), Fe^(2+), Fe^(3+)` and `Au` are x,y,z and w respectively. Then:

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