\(n_{O_2}=\dfrac{6,4}{32}=0,2\left(mol\right)\)

\(n_{CO_2}=\dfrac{13,2}{44}=0,3\left(mol\right)\)

=> \(\overline{M}=\dfrac{6,4+13,2}{0,2+0,3}=39,2\left(g/mol\right)\)

\(a.\)

\(n_{hh}=0.2+0.15+0.1=0.45\left(mol\right)\)

\(V_X=0.45\cdot22.4=10.08\left(l\right)\)

\(b.\)

\(m_X=0.2\cdot28+0.15\cdot71+0.1\cdot32=19.45\left(g\right)\)

\(c.\)

\(\overline{M}_X=\dfrac{19.45}{0.45}=43.22\left(g\text{/}mol\right)\)

\(d.\)

\(d_{X\text{/}kk}=\dfrac{43.22}{29}=1.4\)

Nặng hơn không khí 1.4 lần

\(a.m_{Mg}=0,1.24=2,4\left(g\right)\\ m_{Ca}=0,2.40=8\left(g\right)\\ b.n_{hh}=\dfrac{2,8}{28}+\dfrac{13,2}{44}=0,4\left(mol\right)\\ \Rightarrow V_{hh}=0,4.22,4=8.96\left(l\right)\)

\(a,\left\{{}\begin{matrix}n_{O_2}=1.30\%=0,3\left(mol\right)\\n_{CO_2}=1.20\%=0,2\left(mol\right)\\n_T=1-0,3-0,2=0,5\left(mol\right)\end{matrix}\right.\)

\(b,m_{O_2}=0,3.32=9,6\left(g\right)\)

\(c,m_{hh}=\dfrac{9,6}{49,48\%}=19,4\left(g\right)\\ m_{CO_2}=0,2.44=8,8\left(g\right)\\ \rightarrow m_T=19,4-9,6-8,8=1\left(g\right)\\ \rightarrow M_T=\dfrac{1}{0,5}=2\left(\text{g/mol}\right)\\ \rightarrow T:H_2\)

a. %V (ở cùng điều kiện) cũng là %n

\(Tacó:\%V_T=100-30-20=50\%\\ Trong1molhỗnhợp:\\ n_{O_2}=1.30\%=0,3\left(mol\right)\\ n_{CO_2}=1.20\%=0,2\left(mol\right)\\ n_T=1.50\%=0,5\left(mol\right)\\ b.m_{O_2}=0,3.32=9,6\left(g\right)\\ c.\%m_{O_2}tronghỗnhợplà49,48\%\\ Trong1molhỗnhợp:m_{hh}=\dfrac{9,6}{49,48\%}=19,4\left(g\right)\\ m_{CO_2}=0,2.44=8,8\left(g\right)\\ \Rightarrow m_T=19,4-9,6-8,8=1\left(g\right)\\ \Rightarrow M_T=\dfrac{1}{0,5}=2\\ \Rightarrow TlàH_2\)

Ta có:

$M_{N_2}=28g/mol,\quad M_{O_2}=32g/mol$

Tỉ lệ mol:

$n_{N_2}:n_{O_2}=1:2$

Chọn:

$n_{N_2}=1mol,\quad n_{O_2}=2mol$

Tổng số mol:

$n=1+2=3mol$

Khối lượng hỗn hợp:

$m=1\cdot28+2\cdot32$

$=92g$

Khối lượng mol trung bình:

$M_{tb}=\dfrac{92}{3}\approx30,67g/mol$

$M_{H_2}=2g/mol$

$d_{X/H_2}=\dfrac{M_X}{M_{H_2}}$

$=\dfrac{30,67}{2}\approx15,33$

Kết luận:

$M_{tb}\approx30,67g/mol$

Hỗn hợp khí nặng hơn khí $H_2$ khoảng $15,33$ lần.