\(M_{CH_3COOH}=60\)g/mol

\(\%C=\dfrac{24}{60}\cdot100\%=40\%\)

\(\%H=\dfrac{4}{60}\cdot100\%=6,67\%\)

\(\%O=\dfrac{32}{60}\cdot10\%=53,33\%\)

\(m_C=18\cdot40\%=7,2g\)

\(m_H=\dfrac{1}{15}\cdot18=1,2g\)

\(m_O=18-\left(7,2+1,2\right)=9,6g\)

a/

\(\%C=\dfrac{12.2.100}{60}=40\%\)

\(\%H=\dfrac{1.4.100}{60}=6,7\%\)

\(\%O=100-40-6,7=53,3\%\)

b/

\(n_{CH_3COOH}=\dfrac{18}{60}=0,3mol\)

\(\Rightarrow\left\{{}\begin{matrix}nC=2.0,3=0,6mol\\nH=4.0,3=0,12mol\\nO=2.0,3=0,6mol\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}mC=12.0,6=7,2gam\\mH=1.0,12=0,12gam\\mO=16.0,6=9,6gam\end{matrix}\right.\)

c.

\(n_{nước}=\dfrac{1,8}{18}=0,1mol\)

\(n_H=n_{nước}=0,1mol\)

\(mH=0,1.4=0,4gam\)

\(\%_{Ba}=\dfrac{137}{197}\cdot100\%\approx69,54\%\\ \%_C=\dfrac{12}{197}\cdot100\%\approx6,09\%\\ \%_O=100\%-69,54\%-6,09\%=24,37\%\)

Gọi: \(\left\{{}\begin{matrix}n_{Al_2\left(SO_4\right)_3}=a\left(mol\right)\\n_{Na_2SO_4}=b\left(mol\right)\end{matrix}\right.\) \(\Rightarrow m_X=342a+142b\left(g\right)\)

BTNT O, có: \(n_O=12n_{Al_2\left(SO_4\right)_3}+4n_{Na_2SO_4}=12a+4b\left(mol\right)\)

\(\Rightarrow m_O=16.\left(12a+4b\right)=192a+64b\left(g\right)\)

Mà: O chiếm 50% về khối lượng.

\(\Rightarrow\dfrac{192a+64b}{342a+142b}=0,5\) \(\Rightarrow b=3a\)

\(\Rightarrow\%m_{Al_2\left(SO_4\right)_3}=\dfrac{342a}{342a+142b}.100\%=\dfrac{342a}{342a+142.3a}.100\%=44,53125\%\)

\(\%m_{Na_2SO_4}=100-44,53125=55,46875\%\)

Gọi: \(\left\{{}\begin{matrix}n_{Al_2\left(SO_4\right)_3}=x\left(mol\right)\\n_{CuSO_4}=y\left(mol\right)\end{matrix}\right.\) ⇒ m_hh = 342x + 160y (g)

BTNT O, có: \(n_O=12n_{Al_2\left(SO_4\right)_3}+4n_{CuSO_4}=12x+4y\left(mol\right)\)

⇒ m_O = 16.(12x+4y) = 192x + 64y (g)

Mà: O chiếm 48,34% khối lượng.

\(\Rightarrow\dfrac{192x+64y}{342x+160y}=0,4834\) \(\Rightarrow y=2x\)

\(\Rightarrow\%m_{Al_2\left(SO_4\right)_3}=\dfrac{342x}{342x+160y}.100\%=\dfrac{342x}{342x+160.2x}.100\%\approx51,66\%\)

\(\%m_{CuSO_4}\approx48,34\%\)

\(a.M_{C_2H_6O}=12,2+2+16=46\left(đvC\right)\\ \%C=\dfrac{12.2}{46}.100=52,17\%\\ \%H=\dfrac{6}{46}.100=13,04\%\\ \%O=100-52,17-13,04=34,79\%\\ b.n_{CO_2}=\dfrac{6.6}{44}=0,15\left(mol\right)\\ BTNT\left(C\right):n_{C_2H_6O}.2=n_{CO_2}.1\\ \Rightarrow n_{C_2H_6O}=0,075\left(mol\right)\\ \Rightarrow m_{C_2H_6O}=0,075.46=3,45\left(g\right)\)

\(a,\%m_C=\dfrac{12.2}{12.2+6.1+16}.100\approx52,174\%\\ \%m_H=\dfrac{6.1}{12.2+6.1+16}.100\approx13,043\%\\ \%m_O=\dfrac{16}{12.2+6.1+16}.100\approx34,783\%\)

\(b,n_C=n_{CO_2}=\dfrac{6,6}{44}=0,15\left(mol\right)\\ \Rightarrow m_{C_2H_5OH}=\dfrac{n_C}{2}=\dfrac{0,15}{2}=0,075\left(mol\right)\\ \Rightarrow m_{C_2H_5OH}=0,075.46=3,45\left(g\right)\)

\(a,n_{H_2O}=\dfrac{1,8}{18}=0,1(mol)\\ n_{H_2}=\dfrac{5,6}{22,4}=0,25(mol)\\ n_{Cu}=\dfrac{9.10^{23}}{6.10^{23}}=1,5(mol)\\ m_{O_2}=2.32=64(g)\\ V_{O_2}=2.22,4=44,8(l)\)

a) M_hh = 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) \(\%H=\dfrac{4+3.6}{16+3.30}.100\%=20,75\%\)