Ta có: \(\left\{{}\begin{matrix}n_{NO_2}+n_{CO_2}=0.22\\n_{NO_2}-10n_{CO_2}=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}n_{NO_2}=0.2=n_N\\n_{CO_2}=0.02=n_C\end{matrix}\right.\)

\(P+5HNO_3\rightarrow H_3PO_4+5NO_2+H_2O\)

\(C+4HNO_3\rightarrow4NO_2+CO_2+2H_2O\)

\(S+6HNO_3\rightarrow H_2SO_4+6NO_2+2H_2O\)

Ta có: \(\left\{{}\begin{matrix}12n_C+31n_P+32n_S=0.958\\BTe:4n_C+5n_P+6n_S=0.2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}31n_P+32n_S=0.718\\5n_P+6n_S=0.12\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}n_P=0.018\\n_S=0.005\end{matrix}\right.\)

\(n_{HNO_3}\) ban đầu \(=\dfrac{\left(29\times63\right)}{100\times63}=0.29\) mol

\(BTN:n_{HNO_3}\) dư \(=n_{HNO_3}\) ban đầu \(-n_{NO_2}=0.29-0.2=0.09\) mol

Dung dịch X chứa \(HNO_3\) dư (0.09 mol), \(H_3PO_4\) (0.018 mol) và \(H_2SO_4\) (0.005 mol)

\(n_{H^+}=0.09+0.018\times3+0.005\times2=0.154\) mol

\(n_{KOH}=0.12mol;n_{NaOH}=0.08mol\)

\(\Rightarrow n_{OH^-}=0.12+0.08=0.2mol\)

\(\Rightarrow OH^-\) dư

\(H^++OH^-\rightarrow H_2O\)

0.154               0.154    (mol)

BTKL: \(m_{H_3PO_4}+m_{H_2SO_4}+m_{HNO_3}\) dư \(+m_{KOH}+m_{NaOH}=m_{ct}+m_{H_2O}\)

\(\Leftrightarrow0.018\times98+0.005\times98+0.09\times63+0.12\times56+0.08\times40=m_{ct}+0.154\times18\)

\(\Leftrightarrow m_{ct}=15.072g\)