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\(a.\)
\(GS:\)
\(n_{hh}=1\left(mol\right)\)
\(Đặt:n_{N_2}=a\left(mol\right),n_{CO_2}=b\left(mol\right)\)
\(\Rightarrow a+b=1\left(1\right)\)
\(m_A=28a+44b=18\cdot2=36\left(2\right)\)
\(\left(1\right),\left(2\right):\)
\(a=b=0.5\)
\(\%m_{N_2}=\dfrac{0.5\cdot28}{0.5\cdot28+0.5\cdot44}\cdot100\%=38.89\%\)
\(\%m_{CO_2}=61.11\%\)
\(b.\)
\(\dfrac{n_{N_2}}{n_{CO_2}}=\dfrac{0.5}{0.5}=\dfrac{1}{1}\)
\(n_{N_2}=n_{CO_2}=\dfrac{1}{2}\cdot n_A=\dfrac{0.2}{2}=0.1\left(mol\right)\)
\(Đặt:n_{CO_2}=x\left(mol\right)\)
\(\overline{M}=\dfrac{0.1\cdot28+0.1\cdot44+44x}{0.2+x}=20\cdot2=40\left(\dfrac{g}{mol}\right)\)
\(\Rightarrow x=0.2\)
\(m_{CO_2\left(cầnthêm\right)}=0.2\cdot44=8.8\left(g\right)\)
Câu này anh giải rồi , em xem thử nhé !!
hỗn hợp khí A gồm N2 và CO2 có tỉ khối so với khí Hiđro là 18. Tính thành phần phần trăm khối lượng của mỗi khí trong hợ... - Hoc24
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)`
\(a,\left\{{}\begin{matrix}n_{O_2}=1.30\%=0,3\left(mol\right)\\n_{CO_2}=1.20\%=0,2\left(mol\right)\\n_T=1-0,3-0,2=0,5\left(mol\right)\end{matrix}\right.\)
\(b,m_{O_2}=0,3.32=9,6\left(g\right)\)
\(c,m_{hh}=\dfrac{9,6}{49,48\%}=19,4\left(g\right)\\ m_{CO_2}=0,2.44=8,8\left(g\right)\\ \rightarrow m_T=19,4-9,6-8,8=1\left(g\right)\\ \rightarrow M_T=\dfrac{1}{0,5}=2\left(\text{g/mol}\right)\\ \rightarrow T:H_2\)
a. %V (ở cùng điều kiện) cũng là %n
\(Tacó:\%V_T=100-30-20=50\%\\ Trong1molhỗnhợp:\\ n_{O_2}=1.30\%=0,3\left(mol\right)\\ n_{CO_2}=1.20\%=0,2\left(mol\right)\\ n_T=1.50\%=0,5\left(mol\right)\\ b.m_{O_2}=0,3.32=9,6\left(g\right)\\ c.\%m_{O_2}tronghỗnhợplà49,48\%\\ Trong1molhỗnhợp:m_{hh}=\dfrac{9,6}{49,48\%}=19,4\left(g\right)\\ m_{CO_2}=0,2.44=8,8\left(g\right)\\ \Rightarrow m_T=19,4-9,6-8,8=1\left(g\right)\\ \Rightarrow M_T=\dfrac{1}{0,5}=2\\ \Rightarrow TlàH_2\)
a) Mhh = 13,25.2 = 26,5 (g/mol
Áp dụng sơ đồ đường chéo:
\(\dfrac{V_{CH_4}}{V_{C_2H_6}}=\dfrac{30-26,5}{26,5-16}=\dfrac{1}{3}\\ \rightarrow\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{1}{1+3}.100\%=25\%\\\%V_{C_2H_6}=100\%-25\%=75\%\end{matrix}\right.\)
b) \(\%H=\dfrac{4+3.6}{16+3.30}.100\%=20,75\%\)
a) \(\overline{M}_A=5,875.2=11,75\left(g/mol\right)\)
b) Gọi số mol N2, H2 là a, b (mol)
\(\overline{M}_A=\dfrac{28a+2b}{a+b}=11,75\left(g/mol\right)\)
=> 16,25a = 9,75b
=> a = 0,6b
\(\left\{{}\begin{matrix}\%n_{N_2}=\dfrac{a}{a+b}.100\%=\dfrac{0,6b}{0,6b+b}.100\%=37,5\%\\\%n_{H_2}=\dfrac{b}{a+b}.100\%=\dfrac{b}{0,6b+b}.100\%=62,5\%\end{matrix}\right.\)
c)
1 mol hỗn hợp A chứa \(\left\{{}\begin{matrix}n_{N_2}=\dfrac{1.37,5}{100}=0,375\left(mol\right)\\n_{H_2}=\dfrac{1.62,5}{100}=0,625\left(mol\right)\end{matrix}\right.\)
\(\overline{M}_B=\dfrac{0,375.28+0,625.2+17x}{1+x}=6,4.2=12,8\left(g/mol\right)\)
=> x = 0,25 (mol)