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\(a.\)
\(GS:\)
\(n_{hh}=1\left(mol\right)\)
\(Đặt:n_{N_2}=a\left(mol\right),n_{CO_2}=b\left(mol\right)\)
\(\Rightarrow a+b=1\left(1\right)\)
\(m_A=28a+44b=18\cdot2=36\left(2\right)\)
\(\left(1\right),\left(2\right):\)
\(a=b=0.5\)
\(\%m_{N_2}=\dfrac{0.5\cdot28}{0.5\cdot28+0.5\cdot44}\cdot100\%=38.89\%\)
\(\%m_{CO_2}=61.11\%\)
\(b.\)
\(\dfrac{n_{N_2}}{n_{CO_2}}=\dfrac{0.5}{0.5}=\dfrac{1}{1}\)
\(n_{N_2}=n_{CO_2}=\dfrac{1}{2}\cdot n_A=\dfrac{0.2}{2}=0.1\left(mol\right)\)
\(Đặt:n_{CO_2}=x\left(mol\right)\)
\(\overline{M}=\dfrac{0.1\cdot28+0.1\cdot44+44x}{0.2+x}=20\cdot2=40\left(\dfrac{g}{mol}\right)\)
\(\Rightarrow x=0.2\)
\(m_{CO_2\left(cầnthêm\right)}=0.2\cdot44=8.8\left(g\right)\)
a) Gọi số mol N2, H2 là a, b (mol)
Có: \(\overline{M}_A=\dfrac{28a+2b}{a+b}=7,5.2=15\left(g/mol\right)\)
=> 13a = 13b
=> a = b
=> \(\left\{{}\begin{matrix}\%m_{N_2}=\dfrac{28a}{28a+2b}.100\%=93,33\%\\\%m_{H_2}=\dfrac{2b}{28a+2b}.100\%=6,67\%\end{matrix}\right.\)
b) Giả sử A chứa 1 mol N2, 1 mol H2
PTHH: N2 + 3H2 --to,p,xt--> 2NH3
Xét tỉ lệ: \(\dfrac{1}{1}>\dfrac{1}{3}\) => Hiệu suất tính theo H2
Gọi số mol H2 phản ứng là 3a
PTHH: N2 + 3H2 --to,p,xt--> 2NH3
Trc pư: 1 1 0
Pư: a<--3a--------------->2a
Sau pư: (1-a) (1-3a) 2a
=> \(\overline{M}_B=\dfrac{\left(1-a\right).28+\left(1-3a\right).2+17.2a}{\left(1-a\right)+\left(1-3a\right)+2a}=9,375.2=18,75\left(g/mol\right)\)
=> a = 0,2
=> \(H\%=\dfrac{0,2.3}{1}.100\%=60\%\)
Áp dụng quy tắc đường chéo:
\(a.\\ \Rightarrow\dfrac{V_{Cl_2}}{V_{O_2}}=\dfrac{15,6}{23,4}=\dfrac{2}{3}\\ \Rightarrow\left\{{}\begin{matrix}\%V_{Cl_2}=40\%\\\%V_{O_2}=60\%\end{matrix}\right.\)
\(b.\)
Ta có: \(\dfrac{n_{Cl_2}}{n_{O_2}}=\dfrac{2}{3}\Leftrightarrow\dfrac{m_{Cl_2}}{m_{O_2}}=\dfrac{71.2}{32.3}=\dfrac{71}{48}\Leftrightarrow48m_{Cl_2}-71m_{O_2}=0\)
Mặt khác: \(m_{Cl_2}+m_{O_2}=5,95\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Cl_2}=3,55\left(g\right)\\m_{O_2}=2,4\left(g\right)\end{matrix}\right.\)
Gọi số mol O2, CO2 là a, b
Có: \(\overline{M}=\dfrac{32a+44b}{a+b}=19,5.2=39\)
=> \(a=\dfrac{5}{7}b\)
=> \(\left\{{}\begin{matrix}\%V_{O_2}=\dfrac{a}{a+b}.100\%=\dfrac{\dfrac{5}{7}b}{\dfrac{5}{7}b+b}.100\%=41,67\%\\\%V_{CO_2}=\dfrac{b}{a+b}.100\%=\dfrac{b}{\dfrac{5}{7}b+b}.100\%=58,33\%\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\%m_{O_2}=\dfrac{32a}{32a+44b}.100\%=34,188\%\\\%m_{CO_2}=\dfrac{44b}{32a+44b}.100\%=65,812\%\end{matrix}\right.\)
thay a = \(\dfrac{5}{7}b\) thôi bn :)
\(\%m_{O_2}=\dfrac{32a}{32a+44b}.100\%=\dfrac{32.\dfrac{5}{7}b}{32.\dfrac{5}{7}b+44b}.100\%=34,188\%\)
Giả sử có 1 mol khí Cl2, 2 mol khí O2
a) \(\left\{{}\begin{matrix}\%V_{Cl_2}=\dfrac{1}{1+2}.100\%=33,33\%\\\%V_{O_2}=\dfrac{2}{1+2}.100\%=66,67\%\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\%m_{Cl_2}=\dfrac{1.71}{1.71+2.32}.100\%=52,59\%\\\%m_{O_2}=\dfrac{2.32}{1.71+2.32}.100\%=47,41\%\end{matrix}\right.\)
b) \(\overline{M}=\dfrac{1.71+2.32}{1+2}=45\left(g/mol\right)\)
=> \(d_{A/H_2}=\dfrac{45}{2}=22,5\)
c) \(n_A=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
=> mA = 0,3.45 = 13,5 (g)
a) \(\left\{{}\begin{matrix}n_{Cl_2}+n_{O_2}=\dfrac{6,72}{22,4}=0,3\\\overline{M}=\dfrac{71.n_{Cl_2}+32.n_{O_2}}{n_{Cl_2}+n_{O_2}}=2.29=58\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}n_{Cl_2}=0,2\left(mol\right)\\n_{O_2}=0,1\left(mol\right)\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\%V_{Cl_2}=\dfrac{0,2}{0,3}.100\%=66,67\%\\\%V_{O_2}=\dfrac{0,1}{0,3}.100\%=33,33\%\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}m_{Cl_2}=0,2.71=14,2\left(g\right)\\m_{O_2}=0,1.32=3,2\left(g\right)\end{matrix}\right.\)
Ta có: dA/H2 = 7,5 ⇒ MA = 7,5.2 = 15 (g/mol)
\(\Rightarrow\dfrac{28n_{N_2}+2n_{H_2}}{n_{N_2}+n_{H_2}}=15\)
\(\Rightarrow n_{N_2}=n_{H_2}\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{N_2}=\dfrac{28n_{N_2}}{28n_{N_2}+2n_{H_2}}.100\%=\dfrac{28n_{N_2}}{28n_{N_2}+2n_{N_2}}.100\%\approx93,3\%\\\%m_{H_2}\approx6,7\%\end{matrix}\right.\)