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Ta có: \(n_{H_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)
PTHH:
Mg + 2HCl ---> MgCl2 + H2 (1)
MgO + 2HCl ---> MgCl2 + H2O (2)
Theo PT(1): \(n_{Mg}=n_{H_2}=0,05\left(mol\right)\)
=> \(m_{Mg}=0,05.24=1,2\left(g\right)\)
=> \(\%_{m_{Mg}}=\dfrac{1,2}{9,2}.100\%=13,04\%\)
\(\%_{m_{MgO}}=100\%-13,04\%=86,96\%\)
\(2CH_3COOH+CaCO_3\rightarrow\left(CH_3COO\right)_2Ca+CO_2+H_2O\)
\(n_{CO_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
Theo PT: \(n_{CH_3COOH}=2n_{CO_2}=0,2\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{CH_3COOH}=\dfrac{0,2.60}{21,2}.100\%\approx56,6\%\\\%m_{C_2H_5OH}\approx43,4\%\end{matrix}\right.\)
a, \(n_{Cu\left(OH\right)_2}=\dfrac{4,8}{98}=0,05\left(mol\right)\)
PT: \(2C_3H_5\left(OH\right)_3+Cu\left(OH\right)_2\rightarrow\left[C_3H_5\left(OH\right)_2O\right]_2Cu+H_2O\)
Theo PT: \(n_{C_3H_5\left(OH\right)_3}=2n_{Cu}=0,1\left(mol\right)\)
\(\Rightarrow m_B=21,2-m_{C_3H_5\left(OH\right)_3}=12\left(g\right)\)
Gọi CTPT của B là CnH2n+2O.
PT: \(C_nH_{2n+1}OH+K\rightarrow C_nH_{2n+1}OK+\dfrac{1}{2}H_2\)
\(C_3H_5\left(OH\right)_3+3K\rightarrow C_3H_5\left(OK\right)_3+\dfrac{3}{2}H_2\)
Theo PT: \(n_{H_2}=\dfrac{1}{2}n_{C_nH_{2n+1}OH}+\dfrac{3}{2}n_{C_3H_5\left(OH\right)_3}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
\(\Rightarrow n_{C_nH_{2n+1}OH}=0,2\left(mol\right)\)
\(\Rightarrow M_{C_nH_{2n+2}O}=\dfrac{12}{0,2}=60\left(g/mol\right)\)
\(\Rightarrow14n+18=60\Rightarrow n=3\)
Vậy: B là C3H8O.
\(\Rightarrow\left\{{}\begin{matrix}\%m_{C_3H_5\left(OH\right)_3}=\dfrac{0,1.92}{21,2}.100\%\approx43,4\%\\\%m_{C_3H_8O}\approx56,6\%\end{matrix}\right.\)
b, \(CH_3CH_2CH_2OH+HCl\rightarrow CH_3CH_2CH_2Cl+H_2O\)
\(C_3H_5\left(OH\right)_3+3HCl\rightarrow C_3H_5Cl_3+3H_2O\)
c, BTNT C, có: nCO2 = 3nC3H5(OH)3 + 3nC3H8O = 0,9 (mol)
\(\Rightarrow V_{CO_2}=0,9.22,4=20,16\left(l\right)\)