\(Al\left(x\right)+4HNO_3\rightarrow2H_2O+NO\left(x\right)+Al\left(NO_3\right)_2\)
\(8Al\left(\frac{8y}{3}\right)+30HNO_3\rightarrow15H_2O+3N_2O\left(y\right)+8Al\left(NO_3\right)_2\)
Gọi số mol NO, N2O lần lược là x, y
\(n_{Al}=\frac{4,59}{27}=0,17\)
\(\Rightarrow x+\frac{8y}{3}=0,17\left(1\right)\)
\(M_{hh}=16,75.2=33,5\)
\(\Rightarrow\frac{30x+44y}{x+y}=33,5\)
\(\Leftrightarrow x=3y\left(2\right)\)
Từ (1) và (2) ta có hệ: \(\left\{\begin{matrix}x+\frac{8y}{3}=0,17\\x=3y\end{matrix}\right.\)
\(\Leftrightarrow\left\{\begin{matrix}x=0,09\\y=0,03\end{matrix}\right.\)
\(\Rightarrow V_{NO}=0,09.22,4=2,016\)
\(\Rightarrow V_{N_2O}=0,03.22,4=0,672\)
