a) Gọi số mol Mg, Cu là a,b
=> 24a + 64b = 1,52
\(n_{NO_2}=\dfrac{1,344}{22,4}=0,06\left(mol\right)\)
Mg0 - 2e --> Mg+2
a---->2a---->a
Cu0 - 2e --> Cu+2
b---->2b---->b
N+5 +1e--> N+4
___0,06<--0,06
Bảo toàn e: 2a + 2b = 0,06
=> \(\left\{{}\begin{matrix}a=0,01\\b=0,02\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\%Mg=\dfrac{24.0,01}{1,52}.100\%=15,79\%\\\%Cu=\dfrac{64.0,02}{1,52}.100\%=84,21\%\end{matrix}\right.\)
b) \(m_{Mg\left(NO_3\right)_2}=0,01.148=1,48\left(g\right)\)
\(m_{Cu\left(NO_3\right)_2}=0,02.188=3,76\left(g\right)\)