\(n_{Zn}=\dfrac{23,4}{65}=0,36\left(mol\right);\overline{M_{N_2O,N_2}}=18.2=36\left(g/mol\right)\)Áp dụng sơ đồ đường chéo, ta có: \(\dfrac{n_{N_2}}{n_{N_2O}}=\dfrac{44-36}{36-28}=\dfrac{1}{1}\Rightarrow n_{N_2}=n_{N_2O}=a\left(mol\right)\left(ĐK:a>0\right)\)Quá trình oxi hóa - khử:\(\overset{0}{Zn}\rightarrow\overset{+2}{Zn}+2e\)\(\overset{+5}{2N}+10e\rightarrow\overset{0}{N_2}\)\(\overset{+5}{2N}+8e\rightarrow\overset{+1}{N_2}O\)BTe: \(2n_{Zn}=10n_{N_2}+8n_{N_2O}\)`=>` \(10a+8a=0,36.2=0,72\Leftrightarrow a=0,04\left(mol\right)\)`=>` \(n_{N_2}=n_{N_2O}=0,04\left(mol\right)\)`=>` \(V_{hh.khí}=\left(0,04+0,04\right).22,4=0,896\left(l\right)\)Chọn ACâu trả lời: \(n_{Zn}=\dfrac{23,4}{65}=0,36\left(mol\right);\overline{M_{N_2O,N_2}}=18.2=36\left(g/mol\right)\)Áp dụng sơ đồ đường chéo, ta có: \(\dfrac{n_{N_2}}{n_{N_2O}}=\dfrac{44-36}{36-28}=\dfrac{1}{1}\Rightarrow n_{N_2}=n_{N_2O}=a\left(mol\right)\left(ĐK:a>0\right)\)Quá trình oxi hóa - khử:\(\overset{0}{Zn}\rightarrow\overset{+2}{Zn}+2e\)\(\overset{+5}{2N}+10e\rightarrow\overset{0}{N_2}\)\(\overset{+5}{2N}+8e\rightarrow\overset{+1}{N_2}O\)BTe: \(2n_{Zn}=10n_{N_2}+8n_{N_2O}\)`=>` \(10a+8a=0,36.2=0,72\Leftrightarrow a=0,04\left(mol\right)\)`=>` \(n_{N_2}=n_{N_2O}=0,04\left(mol\right)\)`=>` \(V_{hh.khí}=\left(0,04+0,04\right).22,4=0,896\left(l\right)\)Chọn A