Ta có: \(n_{CaSO_3}=\dfrac{12,1}{120}=\dfrac{121}{1200}\left(mol\right)\)
a. \(PTHH:CaSO_3+H_2SO_4--->CaSO_4+SO_2+H_2O\)
b. Theo PT: \(n_{SO_2}=n_{CaSO_3}=\dfrac{121}{1200}\left(mol\right)\)
\(\Rightarrow V_{SO_2}=\dfrac{121}{1200}.22,4=\dfrac{847}{375}\left(lít\right)\)
c. Theo PT: \(n_{H_2SO_4}=n_{SO_2}=\dfrac{121}{1200}\left(mol\right)\)
\(\Rightarrow C_{M_{H_2SO_4}}=\dfrac{\dfrac{121}{1200}}{\dfrac{300}{1000}}\approx0,336M\)
(Mik nghĩ là đề sai nhé.)