Đáp án B

Gọi nNO, NO₂ lần lượt là 2x và 3x

Bảo toàn n_e=> nAg + 2nCu = 3nNO + nNO_2=> 0,1 + 2.0,04 = 3.2x + 3x => x = 0,02

=> V_X = 5x.22,4 = 2,24 lít => Chọn B.

\(a.\)

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

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

\(b.\)

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

\(c.\)

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

\(d.\)

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

Nặng hơn không khí 1.4 lần

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

Ta có :

\(m_x=m_{CO_2}+m_{O_2}=44.0,15+32.0,1=9,8\) gam

\(n_x=n_{CO_2}+n_{O_2}=0,15+0,1=0,25\) mol

\(\Rightarrow\overline{M_x}=\frac{m_x}{n_x}=\frac{9,8}{0,25}=39,2\) gam/mol

\(\Rightarrow d_{\frac{x}{kk}}=\frac{\overline{M_x}}{29}=\frac{39,2}{29}=1,35\)

\(V_{SO_2}=0,1.22,4=2,24l\)

\(V_{CO}=0,1.22,4=2,24l\\ V_{N_2}=0,1.22,4=2,24l\)

\(\Rightarrow V_{hh}=2,24.3=6,72l\)

Đáp án C