a) Gọi \(\left\{{}\begin{matrix}n_{NO}=a\left(mol\right)\\n_{NO_2}=b\left(mol\right)\end{matrix}\right.\)
Ta lập HPT: \(\left\{{}\begin{matrix}a+b=\dfrac{13,44}{22,4}=0,6\\30a+46b=0,6\cdot1,4023\cdot29\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0,2\\b=0,4\end{matrix}\right.\)
Vì có kim loại Mg nên giả sử p/ứ tạo NH4NO3
Ta có: \(n_{Mg}=\dfrac{14,4}{24}=0,6\left(mol\right)\)
Bảo toàn electron: \(2n_{Mg}=3n_{NO}+n_{NO_2}+8n_{NH_4NO_3}\) \(\Rightarrow n_{NH_4NO_3}=0,025\left(mol\right)\)
\(\Rightarrow m_{muối}=m_{KL}+62\cdot n_{e\left(trao.đổi\right)}+m_{NH_4NO_3}=14,4+0,6\cdot2\cdot62+0,025\cdot80=90,8\left(g\right)\)
b) PT ion: \(Mg^{2+}+2OH^-\rightarrow Mg\left(OH\right)_2\downarrow\)
\(NH_4^++OH^-\rightarrow NH_3\uparrow+H_2O\)
Theo PT: \(n_{OH^-}=2n_{Mg^{2+}}+n_{NH^+_4}=1,225\left(mol\right)=n_{NaOH}\)
\(\Rightarrow V=\dfrac{1,225}{10^{-1}}=12,25\left(l\right)\)