Đáp án:
Giải thích các bước giải:
Bài 9:
`2Na+2H_2O → 2NaOH+H_2↑`
`0,2 → 0,1`
`2K+2H_2O → 2KOH+H_2↑`
`0,1 → 0,05`
a) `n_{Na}=\frac{4,6}{23}=0,2\ mol`
`n_{K}=\frac{3,9}{39}=0,1\ mol`
Theo PT: `n_{H_2}=\frac{1}{2}.n_{Na}=0,1\ mol`
`n_{H_2}=\frac{1}{2}.n_{K}=0,05\ mol`
`⇒ n_{H_2}=0,1+0,05=0,15\ mol`
`V_{H_2}=0,15.22,4=3,36\ \text{lít}`
b) Ta có: `m_{H_2}=0,15.2=0,3\ \text{gam}`
`m_{NaOH}=0,2.40=8\ gam`
`m_{KOH}=0,1.56=5,6\ gam`
Theo định luật bảo toàn khối lượng, ta có:
`m_{dd}=m_{hh}+m_{H_2O}-m_{H_2}`
`m_{dd}=4,6+3,9+91,5-0,3`
`m_{dd}=99,7\ gam`
`C%_{NaOH}=\frac{8}{99,7}.100% \approx 8,02%`
`C%_{KOH}=\frac{5,6}{99,7}.100% \approx 5,62%`
Bài 10:
a) PTHH: `Zn+2HCl \rightarrow ZnCl_2+H_2↑`
b) `n_{Zn}=\frac{97,5}{65}=1,5\ mol`
Theo PT: `n_{H_2}=n_{Zn}=1,5\ mol`
`V_{H_2}=1,5.22,4=33,6\ lít`
