\(m_{Cu}=n_{Cu}\cdot M_{Cu}=0,2\cdot64=12,8\left(g\right)\)
\(m_{Mg}=n_{Mg}\cdot M_{Mg}=0,1\cdot24=2,4\left(g\right)\)
\(m_{Al}=n_{Al}\cdot M_{Al}=0,25\cdot27=6,75\left(g\right)\)
Khối lượng hỗn hợp là:
\(m_{Cu}+m_{Mg}+m_{Al}=12,8+2,4+6,75=21,95\left(g\right)\)