Đáp án:

$V=25,2(l)$

$X:\ C_3H_7O_2N$

Giải thích các bước giải:

 $a,$

$n_{h^2\ khí}=\dfrac{23,52}{22,4}=1,05(mol)$

$n_{N_2}=\dfrac{1,05×1}{6+1}=0,15(mol)$

$n_{CO_2}=1,05-0,15=0,9(mol)$

$n_{H_2O}=\dfrac{18,9}{18}=1,05(mol)$

BTKL: $m_{O_2}=0,15×28+0,9×44+18,9-26,7=26(g)$

$n_{O_2}=\dfrac{36}{32}=1,125(mol)$

$→V=1,125×22,4=25,2(l)$

 $b,$

$\xrightarrow{BTNT\ O}n_{O\ (X)}=0,9×2+1,05-1,125×2=0,6(mol)$

$CT\ X:\ $C_xH_yO_zN_t$

$\xrightarrow{BTNT\ C}n_{C\ (X)}=0,9(mol)$

$\xrightarrow{BTNT\ H}n_{H\ (X)}=1,05×2=2,1(mol)$

$\xrightarrow{BTNT\ N}n_{N\ (X)}=0,15×2=0,3(mol)$

$x:y:z:t=0,9:2,1:0,6:0,3=3:7:2:1$

$→\ CTHH\ X:\ C_3H_7O_2N$

$a)$

Gọi `n_(CO_2)=a,n_(N_2)=b(mol)`

`=>a-6b=0(1)`

`n_(hh)=(23,52)/(22,4)=1,05(mol)`

`=>a+b=1,05(2)`

Từ $(1),(2)$`=>a=0,9;b=0,15`

`m_(CO_2)=0,9.44=39,6(g)`

`m_(N_2)=0,15.28=4,2(g)`

Áp dụng định luật bảo toàn khối lượng :

`m_X+m_(O_2)=m_(N_2)+m_(CO_2)+m_(H_2O)`

`=>m_(O_2)=4,2+39,6+18,9-26,7`

`=>m_(O_2)=36(g)`

`=>n_(O_2)=36/32=1,125(mol)`

`=>V_(O_2)=1,125.22,4=25,2(l)`

$b)$

`n_C=n_(CO_2)=0,9(mol)`

`n_(H_2O)=(18,9)/18=1,05(mol)`

`n_H=2n_(H_2O)=2,1(mol)`

`n_N=0,15.2=0,3(mol)`

`m_C+m_H+m_N`

`=0,9.12+2,1+0,3.14``=17,1(g)`

`=>` Trong $X$ có $C,H,N,O$

`m_O=26,7-17,1=9,6(g)`

`=>n_O=(9,6)/16=0,6(mol)`

Có `n_C:n_H:n_N:n_O`

`=0,9:2,1:0,3:0,6=3:7:1:2`

`=>CTHH` của $X$ là $C_3H_7NO_2$