Câu `1:`

`n_P=\frac{6,2}{31}=0,2(mol)`

`n_{O_2}=\frac{7,84}{22,4}=0,35(mol)`

`a)` `4P+5O_2\overset{t^o}{\to}2P_2O_5`

Do `\frac{0,2}{4}<\frac{0,35}{5}`

`=>` `P` hết, `O_2` dư

`n_{O_2 \text{dư}}=0,35-0,25=0,1(mol)`

`=> m_{O_2 \text{dư}}=0,1.32=3,2g`

`b)` `n_{P_2O_5}=0,5n_P=0,1(mol)`

`=> m_{P_2O_5}=142.0,1=14,2g`

`c)` `n_{O_2 \text{pứ}}=0,25(mol)`

`2KClO_3\overset{t^o}{\to}2KCl+3O_2`

`=> n_{KClO_3}=\frac{2}{3}n_{O_2}=\frac{1}{6}(mol)`

`=> m_{KClO_3}=122,5.\frac{1}{6}\approx 20,42g`

Câu `2:`

`n_{H_2}=\frac{2,24}{22,4}=0,1(mol)`

`a)` `Zn+2HCl\to ZnCl_2+H_2`

`b)` `n_{Zn}=n_{H_2}=0,1(mol)`

`=> a=65.0,1=6,5g`

`c)` `n_{HCl}=2n_{H_2}=0,2(mol)`

`=> V_{HCl}=\frac{m_{\text{dd}}}{n}`

`=> V_{HCl}=\frac{100}{0,2}=500ml =0,5(l)`

`=> CM_{HCl}=\frac{0,2}{0,5}=0,4M`