Giải:Theo bài trong hợp kim \({m_{Ni}} = {\rm{ }}4{m_{Cr}}\)Ta có \({n_{Cr}} = {\rm{ }}1{\rm{ }}mol\)\(\eqalign{& \Rightarrow {m_{Cr}} = {\rm{ }}1.52{\rm{ }} = {\rm{ }}52{\rm{ }}\left( g \right) \cr
& & \Rightarrow {n_{Ni}} = {{208} \over {59}} = 3,53{\rm{ }}\left( {mol} \right) \cr} \)\Rightarrow {m_{Ni}} = {\rm{ }}4.52{\rm{ }} = {\rm{ }}208{\rm{ }}\left( g \right) \cr 