\(CT:H_2SO_4\cdot nSO_3\left(x\right)\)
\(n_{NaOH}=0.1\cdot0.175=0.0175\left(mol\right)\)
\(H^++OH^-\rightarrow H_2O\)
\(0.0175...0.0175\)
\(n_{H_2SO_4}=8.75\cdot10^{-3}\left(mol\right)\)
\(BTS:\)
\(n_{H_2SO_4}=x+xn=8.75\cdot10^{-3}\left(1\right)\)
\(m=98x+80xn=0.826\left(2\right)\)
\(\left(1\right),\left(2\right):\) \(x=7\cdot10^{-3},xn=1.75\cdot10^{-3}\)
\(\Rightarrow n=4\)
\(\dfrac{n_{SO_3}}{n_{H_2SO_4}}=4\)
\(CT:H_2SO_{_{ }4}\cdot4SO_3\)