\(n_{NO}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
\(3Mg+8HNO_3\rightarrow3Mg\left(NO_3\right)_2+2NO+4H_2O\)
x \(\dfrac{8}{3}x\) x \(\dfrac{2}{3}x\) \(\dfrac{4}{3}x\)
\(Fe+4HNO_3\rightarrow Fe\left(NO_3\right)_3+NO+2H_2O\)
y 4y y y 2y
dd Y: \(Mg\left(NO_3\right)_2,Fe\left(NO_3\right)_3\)
gọi x, y là số mol Mg, Fe
có hệ: \(\left\{{}\begin{matrix}24x+56y=9,2\\\dfrac{2}{3}x+y=0,2\end{matrix}\right.\)
<=> x = 0,15 và y = 0,1
a. \(V_{HNO_3}=\dfrac{\left(\dfrac{8}{3}.0,15+4.0,1\right).\left(100+20\right)}{2}:100=0,48\left(l\right)\)
b. \(Mg+4HNO_{3.đn}\rightarrow Mg\left(NO_3\right)_2+2NO_2+2H_2O\)
0,15 0,3
\(Fe+6HNO_{3.đn}\rightarrow Fe\left(NO_3\right)_3+3NO_2+3H_2O\)
0,1 0,3
\(V_{NO_2}=\left(0,3+0,3\right).22,4=13,44\left(l\right)\)