a) `2Na+2HCl->2NaCl+H_2`
`2K+2HCl->2KCl+H_2`
`n_(HCl)=,015.2=0,3(mol)`
`=>n_(Na)+n_(K)=0,3`
Ta có:
`m_(hh)=23n_(Na)+39n_(K)=8,5`
`=>n_(Na)=0,2` và `n_(K)=0,1`
`=>%m_(Na)=(0,2.23)/{8,5} .100=54,12%`
`%m_K=100-54,12=45,88%`
b) `m_(NaCl)=0,2.(23+35,5)=11,7(g)`
`m_(KCl)=0,1.(39+35,5)=7,45(g)`