\(n_{SO_2}=\dfrac{4,48}{22,4}=0,2mol\)

\(\Rightarrow n_{H_2SO_4}=2n_{SO_2}=2\cdot0,2=0,4mol\)

\(\Rightarrow n_{SO_4^{2-}}=n_{H_2SO_4}=0,4mol\Rightarrow m_{SO_4^{2-}}=0,4\cdot96=38,4g\)

\(m=m_{SO_4^{2-}}+m_{hhX}=38,4+6,7=45,1g\)

\(2\left[H\right]+\left[O\right]->H_2O\\ n_{Cl}=n_H=2n_O=\dfrac{44,6-28,6}{16}.2=2mol\\ m_{muoi}=28,6+35,5.2=99,6g\)

\(\left\{{}\begin{matrix}Mg\\Zn\\Fe\end{matrix}\right.+H_2SO_4\rightarrow\left\{{}\begin{matrix}MgSO_4\\ZnSO_4\\FeSO_4\end{matrix}\right.+H_2\uparrow\)

\(m_{SO_4}=86,8-29=57,8\left(g\right)\)

\(\Rightarrow n_{H_2}=n_{H_2SO_4}=n_{SO_4}=0,6\left(mol\right)\)

\(\Rightarrow b=V_{H_2}=0,6.22,4=13,44\left(l\right)\)

Ta có: n_N2 = 0,22 (mol)

⇒ n_NO3^- = 10n_N2 = 2,2 (mol)

⇒ m _muối = m_X + m_NO3^- = 31 + 2,2.62 = 167,4 (g)

n_HNO3 = 12n_N2 = 2,64 (mol)

\(\Rightarrow V_{HNO_3}=\dfrac{2,64}{2}=1,32\left(l\right)=1320\left(ml\right)\)

 Tương tự bài 1, ta có:

m 4 o x i t = m m u o i - m H 2 S O 4 + m H 2 O   s a n   p h a m

m 4 o x i t  = 3,61g

⇒ Chọn C.