Bài 1: CTTQ: XO
PTHH:$XO+2HCl\rightarrow XCl_{2}+H_{2}O$
Ta có:$m_{ctHCl}=\frac{m_{dd}.\frac{C}{100}}{\frac{100}{100}}=\frac{100.21,9}{100}=21,9(g)\\\Rightarrow n_{HCl}=\frac{m}{M}=\frac{21,9}{36,5}=0,6(mol)\\$
Theo pt:$n_{XO}=\frac{1}{2}n_{HCl}=\frac{1}{2.}0,6=0,3(mol)\\\Rightarrow...