Đáp án:
`x=0,54(mol)`
Giải thích các bước giải:
`n_{\text{2,4,6-trinitrophenol }}=\frac{13,74}{229}=0,06(mol)`
`2,4,6-` trinitrophenol $:C_6H_2(NO_2)_3OH \xrightarrow{\text{Thu gọn}}C_6H_3N_3O_7$
$\xrightarrow[]{\text{BTNT C}}:n_{CO_2}+n_{CO}=6.n_{\text{2,4,6-trinitrophenol }}=6.0,06=0,36(mol)$
$\xrightarrow[]{\text{BTNT H}}:n_{H_2}=\frac{3}{2}.n_{\text{2,4,6-trinitrophenol }}=\frac{3}{2}.0,06=0,09(mol)$
$\xrightarrow[]{\text{BTNT N}}:n_{N_2}=\frac{3}{2}.n_{\text{2,4,6-trinitrophenol }}=\frac{3}{2}.0,06=0,09(mol)$
Suy ra `x=n_{CO_2}+n_{CO}+n_{N_2}+n_{H_2}=0,36+0,09+0,09=0,54(mol)`