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

\(m_{hh}=m_{O_2}+m_{N_2}=0.05\cdot32+0.15\cdot28=5.8\left(g\right)\)

\(\overline{M}_X=\dfrac{5.8}{0.05+0.15}=29\left(\dfrac{g}{mol}\right)\)

\(d_{X\text{/}H_2}=\dfrac{29}{2}=14.5\)

\(m_{O2}=n_{O2}.M_{O2}=0,1.32=3,2\left(g\right)\)

\(m_{H2}=n_{H2}.M_{H2}=0,15.2=0,3\left(g\right)\)

\(m_{SO2}=n_{SO2}.M_{SO2}=0,25.64=16\left(g\right)\)

\(m_{hh}=m_{O2}+m_{H2}+m_{SO2}=19,2\left(g\right)\)

P/s: Ko biết lm thế này có đc khum nhỉ ??

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

Chọn B.

$n_X = 0,1 + 0,15 + 0,02 = 0,27(mol)$

$m_X = 0,1.2 + 0,15.30 + 0,02.64 = 5,98(gam)$
$M_X = 5,98 : 0,27 = 22,15(g/mol)$
$d_{X/H_2} = 22,15 : 2 = 11,08$
Đáp án B

Đáp án B nhé bạn 

\(n_{hh}=n_{H_2}+n_{CO_2}+n_{SO_2}=0,1+0,15+0,3=0,55\left(mol\right)\\ \Rightarrow V_{hh\left(đktc\right)}=n.22,4=12,32\left(l\right)\)