Đáp án:
Giải thích các bước giải:
`1)`
`n_{A}=\frac{8,96}{22,4}=0,4(mol)`
`\overline{M_A}=16,75.2=33,5g`/`mol`
Áp dụng quy tắc đường chéo:
`n_{NO}:` `30` `10,5`
`33,5`
`n_{N_2O}:``44` `3,5`
`=> \frac{n_{NO}}{n_{N_2O}}=\frac{10,5}{3,5}=3`
`=> 3n_{N_2O}-n_{NO}=0(mol)(1)`
`n_{N_2O}+n_{NO}=0,4(mol)(2)`
`(1)(2)``=>` $\begin{cases}n_{N_2O}=0,1(mol)\\n_{NO}=0,3(mol)\\\end{cases}$
BTe: `3n_{Al}=8n_{N_2O}+3n_{NO}`
`=> n_{Al}=\frac{8.0,1+0,3.3}{3}=\frac{17}{30}(mol)`
`=> m=\frac{17}{30}.27=15,3g`
`2)`
`n_{Al}=\frac{13,5}{27}=0,5(mol)`
`\overline{M_A}=14,75.2=29,5g`/`mol`
Áp dụng quy tắc đường chéo:
`\frac{n_{NO}}{n_{N_2}}=\frac{30-29,5}{29,5-28}=\frac{1}{3}`
`=> 3n_{NO}=n_{N_2}`
BTe: `3n_{Al}=3n_{NO}+10n_{N_2}`
`=>11n_{N_2}=1,5`
`=> n_{N_2}=\frac{3}{22}(mol)`
`=> n_{NO}=\frac{3}{22}.\frac{1}{3}=\frac{1}{22}(mol)`
`V_{N_2}=\frac{3}{22}.22,4\approx 3,1(l)`
`V_{NO}=\frac{1}{22}.22,4\approx 1,02(l)`
`b)` BT `N`: `n_{HNO_3}=3n_{Al^{3+}}+2n_{N_2}+n_{NO}`
`=> n_{HNO_3}=\frac{20}{11}(mol)`
`=> C_{M\ HNO_3}=\frac{20}{11.2}\approx 0,9M`
