Gọi :
\(\left\{{}\begin{matrix}n_{SO_3}=a\left(mol\right)\\m_{dd\mathbb{H}_2SO_4}=b\left(gam\right)\end{matrix}\right.\)
\(SO_3 + H_2O \to H_2SO_4\)
Ta có :
80a + b = 450
98a + b.49% = 450.83,3%
Suy ra: a = 2,625(mol) ; b = 240 gam
Suy ra: \(m_{SO_3} = 2,625.80 = 210(gam)\)
- Áp dụng pp đường chéo :
m1g dd H2SO4 122,5%.......................34,3
..........................................83,3%
m2g dd H2SO4 49% ...........................39,2
=> \(\left\{{}\begin{matrix}\dfrac{m_1}{m_2}=\dfrac{7}{8}\\m_1+m_2=450\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_1=210\\m_2=240\end{matrix}\right.\) ( g )
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