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`