Ta có: nNO = 0,3 (mol)
\(\Rightarrow n_{NO_3^-\left(trongmuoi\right)}=3n_{NO}=0,9\left(mol\right)\)
\(\Rightarrow m_{muoi}=m_{KL}+m_{NO_3^-\left(trongmuoi\right)}=24,8+0,9.62=80,6\left(g\right)\)
Giả sử: \(\left\{{}\begin{matrix}n_{Fe}=x\left(mol\right)\\n_{Cu}=y\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow56x+64y=24,8\left(1\right)\)
Theo ĐLBT mol e, có: 3x + 2y = 0,3.3
⇒ 3x + 2y = 0,9 (2)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,1\left(mol\right)\\y=0,3\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\frac{0,1.56}{24,8}.100\%\approx22,58\%\\\%m_{Cu}\approx77,42\%\end{matrix}\right.\)
Bạn tham khảo nhé!
