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Tỉ khối của một khí A với khí B là tỉ số về khối lượng mol của khí A so với khí B.
a) \(d_{\dfrac{X}{H_2}}=8\Rightarrow M_X=8.M_{H_2}=8.2.M_H=8.2.1=16\left(\dfrac{g}{mol}\right)\)
b) \(d_{\dfrac{Y}{H_2}}=15\Rightarrow M_Y=15.M_{H_2}=15.2.M_H=15.2.1=30\left(\dfrac{g}{mol}\right)\)
c) \(d_{\dfrac{Z}{H_2}}=32\Rightarrow M_Z=32.M_{H_2}=32.2.M_H=32.2.1=64\left(\dfrac{g}{mol}\right)\)
a) \(M_X=19.2=38\left(g/mol\right)\)
`=>` \(d_{X/kk}=\dfrac{38}{29}=1,310345\)
b) \(m_X=0,4.38=15,2\left(g\right)\)
Gọi \(\left\{{}\begin{matrix}n_{O_2}=x\left(mol\right)\\n_{CO_2}=y\left(mol\right)\end{matrix}\right.\)
`=>` \(\left\{{}\begin{matrix}32x+44y=15,2\\x+y=0,4\end{matrix}\right.\Leftrightarrow x=y=0,2\)
\(m_Y=0,1.28+15,2=18\left(g\right)\)
`=>` \(\left\{{}\begin{matrix}\%m_{N_2}=\dfrac{0,1.28}{18}.100\%=15,56\%\\\%m_{O_2}=\dfrac{0,2.32}{18}.100\%=35,56\%\\\%m_{CO_2}=100\%-15,56\%-35,56\%=48,88\%\end{matrix}\right.\)
b) \(M_{hh}=4.10=40\left(g/mol\right)\)
Gọi \(n_{NO_2}=a\left(mol\right)\)
`=>` \(\left\{{}\begin{matrix}m_{hh}=18+46a\left(g\right)\\n_{hh}=0,5+0,1+a=0,6+a\left(mol\right)\end{matrix}\right.\)
`=>` \(M_{hh}=\dfrac{m_{hh}}{n_{hh}}=\dfrac{18+46a}{0,6+a}=40\)
`=> a = 1`
`=> V_{NO_2(đktc)} = 1.22,4 = 22,4 (l)`
a)
\(\dfrac{M_X}{M_{H_2}}=18\\ \Rightarrow M_X=18.2=32\left(\dfrac{g}{mol}\right)\)
`X:O_2`
b)
\(\dfrac{M_Y}{M_{H_2}}=15\\ \Rightarrow M_Y=15.2=30\left(\dfrac{g}{mol}\right)\)
`Y:NO`
c)
\(\dfrac{M_Z}{M_{H_2}}=32\\ \Rightarrow M_Z=32.2=64\left(\dfrac{g}{mol}\right)\)
`Z:SO_2`
d)
\(\dfrac{M_T}{M_{kk}}=1,517\\ \Rightarrow M_T=1,517.29=44\left(\dfrac{g}{mol}\right)\)
`T:CO_2`
e)
\(\dfrac{M_U}{M_{kk}}=2,759\\ \Rightarrow M_U=2,759.29=80\left(\dfrac{g}{mol}\right)\)
`U:SO_3`
Khối lượng mol phân tử của khí X :
\(d_{X\text{/}H_2}=\frac{M_X}{M_{H_2}}\Rightarrow M_X=d_{X\text{/}H_2}.M_{H_2}=14.2=28\left(g\text{/}mol\right)\)
\(a.\)
\(GS:\)
\(n_{hh}=1\left(mol\right)\)
\(Đặt:n_{N_2}=a\left(mol\right),n_{CO_2}=b\left(mol\right)\)
\(\Rightarrow a+b=1\left(1\right)\)
\(m_A=28a+44b=18\cdot2=36\left(2\right)\)
\(\left(1\right),\left(2\right):\)
\(a=b=0.5\)
\(\%m_{N_2}=\dfrac{0.5\cdot28}{0.5\cdot28+0.5\cdot44}\cdot100\%=38.89\%\)
\(\%m_{CO_2}=61.11\%\)
\(b.\)
\(\dfrac{n_{N_2}}{n_{CO_2}}=\dfrac{0.5}{0.5}=\dfrac{1}{1}\)
\(n_{N_2}=n_{CO_2}=\dfrac{1}{2}\cdot n_A=\dfrac{0.2}{2}=0.1\left(mol\right)\)
\(Đặt:n_{CO_2}=x\left(mol\right)\)
\(\overline{M}=\dfrac{0.1\cdot28+0.1\cdot44+44x}{0.2+x}=20\cdot2=40\left(\dfrac{g}{mol}\right)\)
\(\Rightarrow x=0.2\)
\(m_{CO_2\left(cầnthêm\right)}=0.2\cdot44=8.8\left(g\right)\)
\(d\dfrac{_{M_A}}{M_{H_2}}=32\Rightarrow M_A=32.2=64\left(\dfrac{g}{mol}\right)\)
\(d_{\dfrac{A}{H_2}}=32\\ M_{H_2}=2\left(\dfrac{g}{mol}\right)\\ \Rightarrow M_A=d_{\dfrac{A}{H_2}}.M_{H_2}=32.2=64\left(\dfrac{g}{mol}\right)\)
\(M_X=8\cdot2=16đvC\)
\(M_Y=15\cdot2=30đvC\)
\(M_Z=32\cdot2=64đvC\)