Ta có: \(n_{SO_3}=\dfrac{16}{80}=0,2\left(mol\right)\)
PTHH: SO3 + H2O ---> H2SO4
Theo PT: \(n_{H_2SO_4}=n_{SO_3}=0,2\left(mol\right)\)
Đổi 250ml = 0,25 lít
\(\Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,2}{0,25}=0,8M\)
Chọn B
\(n_{SO_3}=\dfrac{16}{80}=0.2\left(mol\right)\)
\(SO_3+H_2O\rightarrow H_2SO_4\)
\(0.2...................0.2\)
\(C_{M_{H_2SO_4}}=\dfrac{0.2}{0.25}=0.8\left(M\right)\)