\(n_{NO}\)= \(\dfrac{4,48}{22.4}\)= 0,2 (mol)
Ta có PTHH:
1) Al + 4\(HNO_{3_{dư}}\)= 2\(H_2O\) + \(Al\left(NO_3\right)_3\)+ NO↑
27\(x\) (gam) ←--------------------------------- \(x\) (mol)
2) 3Cu + 8\(HNO_{3_{dư}}\)= 4\(H_2O\) + 3\(Cu\left(NO_3\right)_2\)+ 2NO↑
64.3\(y\)(gam) ←------------------------------------ 2\(y\) (mol)
Ta có hệ pt:
\(\left\{{}\begin{matrix}27x+192y=12,3\\x+2y=0,2\end{matrix}\right.\)⇔\(\left\{{}\begin{matrix}x=0,1\\y=0,05\end{matrix}\right.\)
→\(m_{Al}\)= 27.0,1=2,7(gam)
⇒ %Al =\(\dfrac{m_{Al}}{m_{hh}}\) . 100% =\(\dfrac{2.7}{12,3}\).100%=22%
→\(m_{Cu}\)= 192.0,05=9.6(gam)
⇒%Cu= \(\dfrac{9,6}{12,3}\).100%=78%
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