hòa tan alit hơi SO3 ( đo ĐKT)vào 78g H2O thì:
\(SO_3+H_2O--->H_2SO_4\)
\(n_{SO_3}=\dfrac{a}{22,4}\left(mol\right)\)
Theo PTHH: \(n_{H_2SO_4}=\dfrac{a}{22,4}\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4}=4,375a\left(g\right)\)
Ta có: \(C\%_{H_2SO_4}=\dfrac{m_{H_2SO_4}}{m_{ddH_2SO_4}}.100\)
\(\Leftrightarrow25=\dfrac{4,375a}{b}.100\)
\(\Leftrightarrow437,5a-25b=0\)\((I)\)
Ta có: \(m_{ddsau}=m_{SO_3}+m_{H_2O}\)
\(\Leftrightarrow b=4,375a+78\)
\(\Leftrightarrow4,375a-b=-78\)\((II)\)
Từ (I) và (II): \(\left\{{}\begin{matrix}437,5a-25b=0\\4,375a-b=-78\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a=5,94\\b=104\end{matrix}\right.\)
Ta có: \(C_{M_{H_2SO_4}}=\dfrac{10.C\%_{H_2SO_4}.D_{H_2SO_4}}{M_{H_2SO_4}}\)
\(\Leftrightarrow\)\(c=\dfrac{10.25.1,225}{98}=3,125\)
Vậy \(\left\{{}\begin{matrix}a=5,94\\b=104\\c=3,125\end{matrix}\right.\)
