Mình làm mẫu một ý nha:

a, Gọi hh SO₂ và H₂ là X; O₂ và N₂ là Y

\(\Rightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}m_X=0,02.64+0,5.2=2,28\left(g\right)\\n_X=0,02+0,5=0,52\left(mol\right)\end{matrix}\right.\\\left\{{}\begin{matrix}m_Y=0,2.32+0,5.28=20,4\left(g\right)\\n_Y=0,2+0,5=0,7\left(mol\right)\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}M_X=\dfrac{2,28}{0,52}=4,4\left(\dfrac{g}{mol}\right)\\M_Y=\dfrac{20,4}{0,7}=29,1\left(\dfrac{g}{mol}\right)\end{matrix}\right.\)

\(\Rightarrow d_{X/Y}\)\(=\dfrac{4,4}{29,1}=0,15\)

\(a.\)

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

\(b.\)

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

\(c.\)

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

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

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

\(d.\)

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

Vì CO₂ : O₂ = 2 : 1 

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

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

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

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

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

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

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

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

Gọi hh SO₂ và H₂ là X; O₂ và N₂ là Y

mX=0,02.64+0,5.2=2,28(g)

nX=0,02+0,5=0,52(mol)

mY=0,2.32+0,5.28=20,4(g)

nY=0,2+0,5=0,7(mol)

MX=\(\dfrac{2,28}{0,52}\)=4,4(g\mol)

MY=\(\dfrac{20,4}{0,7}\)=29,1(g\mol)

dX/Y=\(\dfrac{4,4}{29,1}=0,15\)

tại sao nữa nha ạ

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

a, \(\overline{M}=\dfrac{0,1.44+0,2.28}{0,1+0,2}\approx33,33\left(g/mol\right)\)

b, \(\overline{M}=\dfrac{0,2.28+0,3.2}{0,2+0,3}=12,4\left(g/mol\right)\)

c, \(\overline{M}=\dfrac{0,1.28+0,2.30+0,2.44}{0,1+0,2+0,2}=35,2\left(g/mol\right)\)

d, \(\overline{M}=\dfrac{0,2.56+0,1.24+0,1.27}{0,2+0,1+0,1}=40,75\left(g/mol\right)\)

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

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

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

a) Theo đề ra: \(\left\{{}\begin{matrix}m_{O2}=0,1\cdot32=3,2\left(gam\right)\\m_{N2}=28\cdot0,25=7\left(gam\right)\\m_{CO}=0,15\cdot28=4,2\left(gam\right)\end{matrix}\right.\)

=> \(\overline{M_{tb}}=\dfrac{m_{O2}+m_{N2}+m_{CO}}{n_{O2}+n_{N2}+n_{CO}}=\dfrac{3,2+7+4,2}{0,1+0,25+0,15}=28,8\left(\dfrac{g}{mol}\right)\) b) d_{hỗn hợp / H2} = \(\dfrac{28,8}{2}=14,4\)

Vậy ....

a) \(\overline{M_{tb}}=\frac{m_{O_2}+m_{N_2}+m_{CO}}{n_{O_2}+n_{N_2}+n_{CO}}\\ =\frac{0,1.32+0,25.28+0,15.28}{0,1+0,25+0,15}\\ =\frac{14,4}{0,5}=28,8\left(\frac{g}{mol}\right)\)

b) \(d_{\frac{hỗnhợp}{KK}}=\frac{28,8}{29}\approx0,993\\ d_{\frac{hỗnhợp}{H_2}}=\frac{28,8}{2}=14,4\)

a) Theo đề ra: ⎧⎪⎨⎪⎩mO2=0,1⋅32=3,2(gam)mN2=28⋅0,25=7(gam)mCO=0,15⋅28=4,2(gam){mO2=0,1⋅32=3,2(gam)mN2=28⋅0,25=7(gam)mCO=0,15⋅28=4,2(gam)

=> 28,82=14,428,82=14,4

Vậy M trung bình là 28.8 ,tỉ khối là 14.4