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Gọi số mol H2S, CO2 là a, b (mol)
\(\overline{M}=\dfrac{34a+44b}{a+b}=19,5.2=39\left(g/mol\right)\)
=> 5a = 5b
=> a = b
=> \(\left\{{}\begin{matrix}\%m_{H_2S}=\dfrac{34a}{34a+44b}.100\%=43,59\%\\\%m_{CO_2}=\dfrac{44b}{34a+44b}.100\%=56,41\%\end{matrix}\right.\)
\(a.\)
\(GS:\)
\(n_{hh}=1\left(mol\right)\)
\(Đặt:n_{N_2}=a\left(mol\right),n_{CO_2}=b\left(mol\right)\)
\(\Rightarrow a+b=1\left(1\right)\)
\(m_A=28a+44b=18\cdot2=36\left(2\right)\)
\(\left(1\right),\left(2\right):\)
\(a=b=0.5\)
\(\%m_{N_2}=\dfrac{0.5\cdot28}{0.5\cdot28+0.5\cdot44}\cdot100\%=38.89\%\)
\(\%m_{CO_2}=61.11\%\)
\(b.\)
\(\dfrac{n_{N_2}}{n_{CO_2}}=\dfrac{0.5}{0.5}=\dfrac{1}{1}\)
\(n_{N_2}=n_{CO_2}=\dfrac{1}{2}\cdot n_A=\dfrac{0.2}{2}=0.1\left(mol\right)\)
\(Đặt:n_{CO_2}=x\left(mol\right)\)
\(\overline{M}=\dfrac{0.1\cdot28+0.1\cdot44+44x}{0.2+x}=20\cdot2=40\left(\dfrac{g}{mol}\right)\)
\(\Rightarrow x=0.2\)
\(m_{CO_2\left(cầnthêm\right)}=0.2\cdot44=8.8\left(g\right)\)
1)
Coi \(n_X = 1(mol)\)
Gọi : \(n_{CO_2} = a(mol) ; n_{N_2} = b(mol)\)
Ta có :
\(n_X = a + b = 1(mol)\\ m_X = 44a + 28b = 1.1,225.32(gam)\\ \Rightarrow a = 0,7 ; b = 0,3\)
Vậy :
\(\%V_{CO_2} = \dfrac{0,7}{1}.100\% = 70\%\\ \%V_{N_2} = 100\% - 70\% = 30\%\)
2)
\(n_X = \dfrac{1}{22,4}(mol)\\ \Rightarrow m_X = n.M = \dfrac{1}{22,4}.1,225.32 = 1,75(gam)\)
\(Gọi\ n_{CO} =a(mol) ; n_{CO_2} = b(mol)\\ n_{khí} = a + b = \dfrac{15,68}{22,4} = 0,7(mol)\\ m_{khí} = 28a + 44b = 27,6(gam)\\ \Rightarrow a = 0,2 ; b = 0,5\\ \%m_{CO} = \dfrac{0,2.28}{27,6}.100\% = 20,29\%\\ \%m_{CO_2} = 100\% - 20,29\% = 79,71\%\)
cho mik hỏi bạn tính sao để có thể ra đc 0,2 vạy
\(C_3H_4+4O_2\underrightarrow{^{^{t^0}}}3CO_2+2H_2O\)
\(C_3H_6+\dfrac{9}{2}O_2\underrightarrow{^{^{t^0}}}3CO_2+3H_2O\)
\(C_3H_8+5O_2\underrightarrow{^{^{t^0}}}3CO_2+4H_2O\)
\(n_X=\dfrac{1.12}{22.4}=0.05\left(mol\right)\)
\(\Rightarrow n_{CO_2}=3\cdot n_X=3\cdot0.05=0.15\left(mol\right)\)
\(m_{CO_2}=0.15\cdot44=6.6\left(g\right)\)
\(m_X=21\cdot2\cdot0.05=2.1\left(g\right)\)
\(\Rightarrow m_H=m_X-m_C=2.1-0.15\cdot12=0.3\left(g\right)\)
\(n_H=0.3=0.3\left(mol\right)\)
\(\Rightarrow n_{H_2O}=0.15\left(mol\right)\)
\(m_{H_2O}=0.15\cdot18=2.7\left(g\right)\)
Gọi công thức hóa học chung của hỗn hợp X là \(C_3H_x\)
Có \(\overline{M_X}=21\cdot2=42đvC\Rightarrow12\cdot3+x\cdot1=42\)
\(\Rightarrow x=6\)
\(n_X=\dfrac{1,12}{22,4}=0,05mol\)
\(C_3H_6+\dfrac{9}{2}O_2\rightarrow3CO_2+3H_2O\)
0,05 0,15 0,15
\(m_{CO_2}=0,15\cdot44=6,6g\)
\(m_{H_2O}=0,15\cdot18=2,7g\)
1)
Coi nX=1(mol)nX=1(mol)
Gọi : nCO2=a(mol);nN2=b(mol)nCO2=a(mol);nN2=b(mol)
Ta có :
nX=a+b=1(mol)mX=44a+28b=1.1,225.32(gam)⇒a=0,7;b=0,3nX=a+b=1(mol)mX=44a+28b=1.1,225.32(gam)⇒a=0,7;b=0,3
Vậy :
%VCO2=0,71.100%=70%%VN2=100%−70%=30%