Bài `9`:
`n_{Cu}=\frac{6,4}{64}=0,1(mol)`
`a)` Phương trình:
`CuO+H_2\overset{t^o}{\to}Cu+H_2O`
`n_{H_2}=n_{Cu}=0,1(mol)`
`=> V_{H_2}=0,1.22,4=2,24(l)`
`b)` `n_{H_2}=0,1(mol)`
`2Na+2H_2O\to 2NaOH+H_2`
`n_{H_2}.2=n_{Na}=n_{NaOH}=0,2(mol)`
`=> m=23.0,2=4,6g`
`c)` `m_{\text{dd spu}}=m_{Na}+m_{H_2O}-m_{H_2}`
`=> m_{\text{dd spu}}=4,6+500-0,1.2=504,4g`
`=> C%_{NaOH}=\frac{0,2.40.100%}{504,4}\approx 1,59%`