a) Ta có: nhh A = \(\dfrac{3,136}{22,4}=0,14\left(mol\right)\)
hay: n\(N_2\) + n\(H_2\) = 0,14(mol)
mà: nN2 : nH2 = 3:4
=>\(\left\{{}\begin{matrix}n_{N_2}=0,06\left(mol\right)\_{H_2}=0,08\left(mol\right)\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}V_{N_2}=0,06.22,4=1,344\left(l\right)\\V_{H_2}=0,08.22,4=1,792\left(l\right)\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\%V_{N_2}=\dfrac{1,344}{3,136}.100\%\simeq42,86\%\\\%V_{H_2}=\dfrac{1,792}{3,136}.100\%\simeq57,14\%\end{matrix}\right.\)
Lại có: \(\left\{{}\begin{matrix}m_{N_2}=0,06.28=1,68\left(g\right)\\m_{H_2}=0,08.2=0,16\left(g\right)\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\%m_{N_2}=\dfrac{1,68}{1,68+0,16}.100\%\simeq91,3\%\\\%m_{H_2}=\dfrac{0,16}{1,68+0,16}.100\%\simeq8,7\%\end{matrix}\right.\)
b) Ta có: Mhh A = \(\dfrac{m_{hhA}}{n_{hhA}}=\dfrac{1,84}{0,06+0,08}\simeq13,14\left(g\right)\)
mà: Mkk = 29(g)
nên: dhh A/ kk =\(\dfrac{13,14}{29}\simeq0,453\)
Vậy hỗn hợp A nhẹ hơn không khí 0,453 lần.
