14. Đặt \(n_{Al}=n_{Al_2O_3}=a\left(mol\right)\)
`=> 27a + 102a = 15,48 <=> a = 0,12 (mol)`
BTNT Al: \(n_{Al\left(NO_3\right)_3}=n_{Al}+2n_{Al_2O_3}=0,36\left(mol\right)\)
`=>` \(n_{NH_4NO_3}=\dfrac{77,88-0,36.213}{80}=0,015\left(mol\right)\)
Quá trình oxi hóa - khử:
\(\overset{0}{Al}\rightarrow\overset{+3}{Al}+3e\\ \overset{+5}{2N}+8e\rightarrow\overset{+1}{N_2}O\\ \overset{+5}{N}+8e\rightarrow\overset{-3}{N}H_4NO_3\)
BTe: \(3n_{Al}=8n_{N_2O}+8n_{NH_4NO_3}\)
`=>` \(n_{N_2O}=\dfrac{3.0,12-8.0,015}{8}=0,03\left(mol\right)\)
`=> V = V_{N_2O} = 0,03.22,4 = 0,672 (l)`
BTNT N: \(y=n_{HNO_3}=3n_{Al\left(NO_3\right)_3}+2n_{NH_4NO_3}+2n_{N_2O}=1,17\left(mol\right)\)
