`n_(NaOH) = 20 / 40 = 0.5 ( mol ) `
`PT : CuCl_2 + 2NaOH -> Cu(OH)_2 ↓ + 2NaCl`
đề : ` 0.2 - 0.5`
pứ : ` 0.2 - 0.4 - 0.2 - 0.4`
hết : ` 0 - 0.1 - 0.2 - 0.4`
* Kết tủa là : `Cu(OH)_2`
* Nung thui :
`Cu(OH)_2 \overset{t^o}to\ CuO + H_2O`
` 0.2 - 0.2`
b) `m_(CuO) = 0.2 * 80 = 16 ( g ) `
c)` m_(NaOH)dư = 0.1 * 40 = 4 ( g )`
`m_(NaCl) = 0.4 * 58.5 = 23.4 ( g )`