$n_{hỗn\ hợp\ khí} = 0,25 + 0,15 = 0,4(mol)$
$V_{hỗn\ hợp\ khí} = 0,4.22,4 = 8,96(lít)$

Câu 1: \(m_{hh}=0,2\cdot27+0,4\cdot39=21\left(g\right)\)

Câu 2: \(V_{khí}=\left(0,1+0,15\right)\cdot22,4=5,6\left(l\right)\)

Câu 3: \(n_{NO_2}=\dfrac{6,9}{46}=0,15\left(mol\right)\) \(\Rightarrow V_{khí}=\left(0,15+0,15\right)\cdot22,4=6,72\left(l\right)\)

Câu 4: 

Ta có: \(\left\{{}\begin{matrix}n_{CO_2}=\dfrac{0,672}{22,4}=0,03\left(mol\right)\\n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\end{matrix}\right.\) \(\Rightarrow m_{hh}=0,03\cdot44+0,4\cdot2=2,12\left(l\right)\)

\(a.\)

\(m_{hh}=0.12\cdot90+0.15\cdot58=19.5\left(g\right)\)

\(b.\)

\(V_{hh}=\left(0.25+0.1+0.05\right)\cdot22.4=8.96\left(l\right)\)

\(c.\)

\(n_A=\dfrac{10.08}{22.4}=0.45\left(mol\right)\)

\(M_A=23\cdot2=46\left(\dfrac{g}{mol}\right)\)

\(m_A=0.45\cdot46=20.7\left(g\right)\)

\(d.\)

\(n_{hh}=\dfrac{6.72}{22.4}=0.3\left(mol\right)\)

Vì CO₂ : O₂ = 2 : 1 

\(\Rightarrow n_{CO_2}=0.2\left(mol\right),n_{O_2}=0.1\left(mol\right)\)

\(m_{hh}=0.2\cdot44+0.1\cdot32=12\left(g\right)\)

\(\overline{M}=\dfrac{12}{0.3}=40\left(\dfrac{g}{mol}\right)\)

m h h = m C O 2 + m O 2 = (0,3.44) + (0,15.32) = 18(g)

→ V C O  = 22,4. n C O 2 + n O 2 = 10,08(l)

\(a.\)

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

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

\(b.\)

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

\(c.\)

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

\(d.\)

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

Nặng hơn không khí 1.4 lần

a)

b) VCO2 = 22,4 .0,175 = 3,92l.

VH2 = 22,4 .1,25 = 28l.

VN2 = 22,4.3 = 67,2l.

c) Số mol của hỗn hợp khí bằng tổng số mol của từng khí.

nhh = nCO2 + nH2 + nN2 = 0,01 + 0,02 + 0,02 = 0,05 mol

Vhh khí = (0,01 + 0,02 + 0,02) . 22,4 = 1,12l.

\(n_{O_2}=\dfrac{3,2}{32}=0,1\left(mol\right)\)

=> V_hh = (0,1+0,15).22,4 = 5,6 (l)