Đáp án:
Giải thích các bước giải:
Bài `10`
`a.`
`n_{SO_2}=\frac{V}{22,4}=\frac{5,6}{22,4}=0,25(mol)`
`N_{SO_2}=n.6,10^23=0,25.6.10^23=1,5.10^23`(phân tử)
`m_{SO_2}=m.M=0,25.64=16(g)`
`b.`
`n_{N_2}=\frac{V}{22,4}=\frac{8,96}{22,4}=0,4(mol)`
`N_{N_2}=n.6,10^23=0,4.6.10^23=2,4.10^23`(phân tử)
`m_{N_2}=n.M=0,4.2811,2(g)`
`c.`
`n_{C_4H_10}=\frac{V}{22,4}=\frac{11,2}{22,4}=0,5(mol)`
`N_{C_4H_10}=n.6,10^23=0,5.6.10^23=3.10^23`(phân tử)
`m_{C_4H_10}=m.M=0,5.58=29(g)`
`d.`
`n_{H_2S}=\frac{V}{22,4}=\frac{16,8}{22,4}=0,75(mol)`
`N_{H_2S}=n.6,10^23=0,65.6.10^23=4,5.10^23`(phân tử)
`m_{H_2S}=m.M=0,75.34=25,5(g)`
Bài `11`
`a.`
`n_{SO_3}=\frac{V}{22,4}=\frac{11,2}{22,4}=0,5(mol)`
`N_{SO_3}=n.6,10^23=0,5.6.10^23=3.10^23`(phân tử)
`m_{SO_3}=m.M=0,5.80=40(g)`
`b.`
`n_{NH_3}=\frac{V}{22,4}=\frac{8,4}{22,4}=0,375(mol)`
`N_{NH_3}=n.6,10^23=0,375.6.10^23=2,25.10^23`(phân tử)
`m_{NH_3}=m.M=0,375.17=6,375(g)`
`c.`
`n_{CH_4}=\frac{V}{22,4}=\frac{5,6}{22,4}=0,25(mol)`
`N_{CH_4}=n.6,10^23=0,25.6.10^23=1,5.10^23`(phân tử)
`m_{CH_4}=m.M=0,25.16=4(g)`
`d.`
`n_{SO_3}=\frac{V}{22,4}=\frac{8,4}{22,4}=0,375(mol)`
`N_{SO_3}=n.6,10^23=0,375.6.10^23=2,25.10^23`(phân tử)
`m_{SO_3}=m.M=0,375.80=30(g)`
`#Devil`
