Đáp án:
Giải thích các bước giải:
`a.`
`2Na+2H_2O->2NaOH+H_2↑`
`2K+2H_2O->2KOH+H_2↑`
`n_{H_2}=(2,24)/(22,4)=0,1(mol)`
`BTKL:`
`m_A+m_{H_2O}=m_B+m_{H_2}`
`=>6,2+94=m_B+0,1.2`
`=>m_B=6,2+94-0,2=100(g)`
`b.`
Gọi `a,b` là `n_{Na},n_K`
`m_{hh}=23a+39b=6,2(1)`
`∑n_{H_2}=0,5a+0,5b=0,1(2)`
`(1),(2)=>a=0,1;b=0,1`
`=>`
`n_{Na}=a=0,1(mol)`
`n_K=b=0,1(mol)`
`=>`
`n_{NaOH}=n_{Na}=0,1(mol)`
`n_{KOH}=n_K=0,1(mol)`
`=>`
`m_{NaOH}=0,1.40=4(g)`
`m_{KOH}=0,1.56=5,6(g)`
`=>`
`C%_{NaOH}=4/100 .100=4%`
`C%_{KOH}=(5,6)/100 .100=5,6%`