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\(n_{CH_3COOH}=\dfrac{12}{60}=0,2\left(mol\right)\)
PTHH: CH3COOH + MHCO3 --> CH3COOM + CO2 + H2O
0,2------->0,2--------------------->0,2
=> VCO2 = 0,2.22,4 = 4,48 (l)
\(m_{MHCO_3}=\dfrac{200.10}{100}=20\left(g\right)\)
=> \(M_{MHCO_3}=\dfrac{20}{0,2}=100\left(g/mol\right)\)
=> MM = 39 (g/mol)
=> M là K
a, \(2K+2CH_3COOH\rightarrow2CH_3COOK+H_2\)
b, \(C_2H_5OH+O_2\underrightarrow{t^o,xt}CH_3COOH+H_2O\)
Ta có: \(n_{CH_3COOH}=\dfrac{12}{60}=0,2\left(mol\right)\)
Theo PT: \(n_{C_2H_5OH\left(LT\right)}=n_{CH_3COOH}=0,2\left(mol\right)\)
Mà: H = 80% \(\Rightarrow n_{C_2H_5OH\left(TT\right)}=\dfrac{0,2}{80\%}=0,25\left(mol\right)\)
\(\Rightarrow m_{C_2H_5OH\left(TT\right)}=0,25.46=11,5\left(g\right)\)
\(\Rightarrow V_{C_2H_5OH\left(TT\right)}=\dfrac{11,5}{0,8}=14,375\left(ml\right)\)
\(\Rightarrow V_{C_2H_5OH\left(10^o\right)}=\dfrac{14,375}{10}.100=143,75\left(ml\right)\)
\(V_{C_2H_5OH}=\dfrac{1.1000.40}{100}=400\left(ml\right)\\ \rightarrow m_{C_2H_5OH\left(TT\right)}=400.0,8=320\left(g\right)\\ \rightarrow m_{C_2H_5OH\left(LT\right)}=\dfrac{320.100}{80}=400\left(g\right)\\ \rightarrow n_{C_2H_5OH\left(LT\right)}=\dfrac{400}{23}\left(mol\right)\)
PTHH: \(C_6H_{12}O_6\underrightarrow{\text{men rượu}}2C_2H_5OH+2CO_2\uparrow\)
\(\dfrac{200}{23}\)<-----------------\(\dfrac{400}{23}\)
\(\rightarrow m_{C_6H_{12}O_6}=\dfrac{200}{23}.180=\dfrac{36000}{23}\left(g\right)\)
\(m_{CH_3COOH}=150.30\%=45\left(g\right)\\ n_{CH_3COOH}=\dfrac{45}{60}=0,75\left(mol\right)\)
PTHH: CH3COOH + C2H5OH \(\xrightarrow[t^o]{H_2SO_4đặc}\) CH3COOC2H5 + H2O
0,75 ----------> 0,75 ------------------> 0,75
\(V_{C_2H_5OH}=\dfrac{46.0,75}{0,8}=43,125\left(ml\right)\\ m_{CH_3COOC_2H_5}=0,75.45\%.88=29,7\left(g\right)\)