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Bài 1
C+O2---->CO2
CO2 + NaOH---->NaHCO3
2NaHCO3---->Na2CO3+H2O+CO2
Na2CO3+CaCl2--->2NaCl+CaCO3
CaCO3---->CaO+CO2
CO2+C--->2CO
2CO+O2--->2CO2
CO2+H2O--->H2CO3
Bài 2
\(CH3COOH+NaOH-->CH3COONa+H2O\)
\(n_{CH3COOH}=0,4.2=0,8\left(mol\right)\)
\(n_{NaOH}=n_{CH3COOH}=0,8.40=32\left(g\right)\)
\(m_{dd}=\frac{32.100}{12}=266,67\left(g\right)\)
\(a,C_2H_6O+3O_2\xrightarrow[]{t^o}2CO_2+3H_2O\\ CO_2+2NaOH\rightarrow Na_2CO_3+H_2O\\ Na_2CO_3+CO_2+H_2O\rightarrow2NaHCO_3\\ NaHCO_3+HCl\rightarrow NaCl+CO_2\uparrow+H_2O\)
\(b,C_2H_4+H_2O\xrightarrow[axit]{t^o}C_2H_5OH\\ C_2H_5OH+O_2\underrightarrow{\text{men giấm}}CH_3COOH+H_2O\\ CH_3COOH+C_2H_5OH\xrightarrow[t^o]{H_2SO_{4\left(đặc\right)}}CH_3COOC_2H_5+H_2O\\ 2CH_3COOH+BaO\rightarrow\left(CH_3COO\right)_2Ba+H_2O\\ \left(CH_3COO\right)_2Ba+Na_2SO_4\rightarrow2CH_3COONa+BaSO_4\downarrow\\ CH_3COONa+NaOH\rightarrow Na_2CO_3+CH_4\uparrow\)
– Công thức cấu tạo của axit acrylic là CH2=CH–COOH
– Các phương trình phản ứng:
CH2=CH–COOH + H2 →CH3–CH2–COOH
CH2=CH–COOH + Br2 →CH2Br–CHBr–COOH
2CH2=CH–COOH + 2Na →2CH2=CH–COONa + H2
CH2=CH–COOH + NaOH → CH2=CH–COONa + H2O
2CH2=CH–COOH + Na2CO3 →2CH2=CH–COONa + H2O + CO2
CH2=CH–COOH + C2H5OH →CH2=CH–COOC2H5 + H2O
$a\big)$
$n_{CH_3COOH}=\dfrac{100}{1000}.1=0,1(mol)$
$CH_3COOH+NaOH\to CH_3COONa+H_2O$
Theo PT: $n_{NaOH}=n_{CH_3COOH}=0,1(mol)$
$\to C\%_{NaOH}=\dfrac{0,1.40}{50}.100\%=80\%$
$b\big)$
$n_{Na_2CO_3}=\dfrac{10,6}{106}=0,1(mol)$
$2CH_3COOH+Na_2CO_3\to 2CH_3COONa+CO_2+H_2O$
Theo PT: $\begin{cases} n_{CO_2}=n_{Na_2CO_3}=0,1(mol)\\ n_{CH_3COONa}=2n_{Na_2CO_3}=0,2(mol) \end{cases}$
$\to C\%_{CH_3COONa}=\dfrac{0,2.82}{60+10,6-0,1.44}.100\%\approx 24,77\%$
nKOH = 0,5.0,3 = 0,15 mol
CH3COOH + KOH → CH3COOK + H2O
0,15 0,15 0,15 mol
a) CM CH3COOH = 0,15/0,2 =0,75M
b) Thể tích của dung dịch thu được sau phản ứng: 500 ml
CM CH3COOK = 0,15/0,5 = 0,3M
c) Phản ứng lên men giấm
C2H5OH + O2 → CH3COOH + H2O
0,15 0,15
→ mC2H5OH = 0,15.46 = 6,9 gam
\(n_{KOH}=0,5\cdot0,3=0,15mol\)
\(CH_3COOH+KOH\rightarrow CH_3COOK+H_2O\)
0,15 0,15 0,15 0,15
a)\(C_{M_{CH_3COOH}}=\dfrac{0,15}{0,2}=0,75M\)
b)\(C_{M_{CH_3COOK}}=\dfrac{0,15}{0,2+0,3}=0,3M\)
\(n_{CH_3COOH}=\dfrac{130.12}{100.60}=0,26\left(mol\right)\)
\(CaCO_3+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Ca+CO_2+H_2O\)
0,13 0,26 0,13 0,13 ( mol )
\(m_{CaCO_3}=0,13.100=13\left(g\right)\)
\(V_{CO_2}=0,13.22,4=2,912\left(l\right)\)
\(m_{ddspứ}=13+130-0,13.44=137,28\left(g\right)\)
\(C\%_{\left(CH_3COO\right)_3Ca}=\dfrac{0,13.158}{137,28}.100=14,96\%\)
1,
\(C+O_2\rightarrow CO_2\)
\(CO_2+NaOH\rightarrow NaHCO_3\)
\(NaHCO_3+NaOH\rightarrow Na_2CO_3+H_2O\)
\(Na_2CO_3+CaCl_2+2NaCl+CaCO_3\)
\(CaCO_3\rightarrow CaO+CO_2\)
\(CO_2+C\rightarrow2CO\)
\(2CO+O_2\rightarrow2CO_2\)
\(CO_2+H_2O\rightarrow H_2CO_3\)
2,
\(CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O\)
Ta có :
\(n_{NaOH}=n_{CH3COOH}=0,4.2=0,8\left(mol\right)\)
\(\Rightarrow m_{dd_{NaOH}}=\frac{0,8.40}{12\%}=266,67\left(g\right)\)
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