`2NaOH+H_2SO_4->Na_2SO_4+2H_2O`
`2KOH+H_2SO_4->K_2SO_4+2H_2O`
Gọi `a,b` là `n_{NaOH},n_{KOH}`
`=>`
`m_{hh}=40a+56b=3,04(1)`
`m_{\text{muối}}=142.0,5a+174.0,5b=71a+87b=4,9(2)`
`(1),(2)=>a=0,02;b=0,04`
`m_{NaOH}=0,02.40=0,8(g)`
`m_{KOH}=0,04.56=2,24(g)`
`\text{___________________________________}`
Đáp án:
`m_{NaOH}=0,8(g)`
`m_{KOH}=2,24(g)`