\(n_{SO_3}=a\left(mol\right)\)
\(m_{H_2SO_4}=500\cdot22.5\%=112.5\left(g\right)\)
\(SO_3+H_2O\rightarrow H_2SO_4\)
\(a.................a\)
\(m_{H_2SO_4}=98a+112.5\left(g\right)\)
\(m_{dd_{H_2SO_4}}=80a+500\left(g\right)\)
\(C\%_{H_2SO_4}=\dfrac{98a+112.5}{80a+500}\cdot100\%=42.5\%\)
\(\Leftrightarrow a=1.5625\)
\(m_{SO_3}=125\left(g\right)\)
Số mol SO3 cần thân vào là x (mol)
Ta có: SO3 + H2O -> H2SO4
x x
mH2SO4 tạo thành là: 98x (gam)
mH2SO4 22,5% là: 500 x 22,5% = 112,5 (gam)
-> (98x + 112,5). 100/80x + 500 = 42,5
-> x = 1,5625
mSO3 = 1,5625 x 80 = 125 (gam)
