+

`n_{CH_4}=m/M={1,6}/16=0,1(mol)`

`V_{CH_4(dktc)}=n.22,4=0,1.22,4=2,24(l)`

+

`n_{NH_3}=n/M={3,4}/17=0,2(mol)`

`V_{NH_3(dktc)}=n.22,4=0,2.22,4=4,48(l)`

+

`n_{N_2O}={3.10^{23}}/{6.10^{23}}=0,5(mol)`

`V_{N_2O(dktc)}=n.22,4=0,5.22,4=11,2(l)`

Đáp án:

`-V_{CH_4}(đktc) =2,24(l)`

`-V_{NH_3}(đktc) =4,48(l)`

`-V_{N_2O}(đktc) =11,2(l)`

Giải thích các bước giải:

`-n_{CH_4}=\frac{m_{CH_4}}{M_{CH_4}}=\frac{1,6}{16}=0,1(mol)`

`→V_{CH_4}(đktc)=n_{CH_4}.22,4=0,1.22,4=2,24(l)`

`-n_{NH_3}=\frac{m_{NH_3}}{M_{NH_3}}=\frac{3,4}{17}=0,2(mol)`

`→V_{NH_3}(đktc)=n_{NH_3}.22,4=0,2.22,4=4,48(l)`

- Số Avogađro `1(mol)≈6.10^{23}` phân tử 

`→n_{N_2O}=\frac{3.10^{23}}{6.10^{23}}=0,5(mol)`

`→V_{N_2O}(đktc)=n_{N_2O}.22,4=0,5.22,4=11,2(l)`