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\(n_{C_2H_5OH}=\dfrac{9,2}{46}=0,2\left(mol\right)\)
\(C_2H_5OH+3O_2\underrightarrow{t^o}2CO_2+2H_2O\)
0,2 0,6 0,4 0,4
\(a,V_{O_2}=0,6.22,4=13,44\left(l\right)\)
\(V_{kk}=13,44.5=67,2\left(l\right)\)
b, \(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3\downarrow+H_2O\)
0,4 0,4
\(m_{CaCO_3}=0,4.100=40\left(g\right)\)
\(m_{CaCO_3tt}=40.95\%=38\left(g\right)\)
Câu 1.
\(V_{C_2H_5OH}=\dfrac{90.90}{100}=81\left(ml\right)\)
\(m_{C_2H_5OH}=81.0,8=64,8g\)
\(n_{C_2H_5OH}=\dfrac{64,8}{46}=1,4mol\)
\(2C_2H_5OH+2Na\rightarrow2C_2H_5ONa+H_2\)
1,4 0,7 ( mol )
\(V_{H_2}=0,7.22,4=15,68l\)
Câu 2.
\(n_{CaCO_3}=\dfrac{100}{100}=1mol\)
\(Ca\left(OH\right)_2+CO_2\rightarrow CaCO_3+H_2O\)
1 1 ( mol )
\(C_2H_5OH+3O_2\rightarrow\left(t^o\right)2CO_2+3H_2O\)
0,5 1,5 1 ( mol )
\(V_{kk}=\left(1,5.22,4\right).5=168l\)
\(m_{C_2H_5OH}=0,5.46=23g\)
\(V_{C_2H_5OH}=\dfrac{23}{0,8}=28,75ml\)
Độ rượu = \(\dfrac{28,75}{30}.100=95,83^o\)
A tác dụng với NaHCO3 cho khí CO2 → A: axit CH3COOH
BTKL: m + mO2 = mCO2 + mH2O => m = 1,8
=> nCH3COOH = 0,03
CH3COOH + C2H5OH → CH3COOC2H5 + H2O
0,03 0,02 0,0125
=> H = 62,5%
\(n_{CH_4}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
PTHH: CH4 + 2O2 --to--> CO2 + 2H2O
0,2 0,4 0,2
CO2 + Ca(OH)2 ---> CaCO3 + H2O
0,2 0,2
\(\rightarrow\left\{{}\begin{matrix}x=0,2.100=20\left(g\right)\\V_{kk}=0,4.5.22,4=44,8\left(l\right)\end{matrix}\right.\)
a, \(n_{C_2H_4}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)
PT: \(C_2H_4+3O_2\underrightarrow{t^o}2CO_2+2H_2O\)
\(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_{3\downarrow}+H_2O\)
Theo PT: \(n_{CaCO_3}=n_{CO_2}=2n_{C_2H_4}=0,1\left(mol\right)\)
\(\Rightarrow m_{CaCO_3}=0,1.100=10\left(g\right)\)
b, Theo PT: \(n_{O_2}=3n_{C_2H_4}=0,15\left(mol\right)\Rightarrow V_{O_2}=0,15.22,4=3,36\left(l\right)\)
\(\Rightarrow V_{kk}=\dfrac{V_{O_2}}{20\%}=16,8\left(l\right)\)
\(M_X=\dfrac{1.5}{\dfrac{1.775}{71}}=60\left(g\text{/}mol\right)\)
\(n_X=\dfrac{2.25}{60}=0.0375\left(mol\right)\)
CT : \(C_xH_yO_z\)
\(n_{CaCO_3}=n_{CO_2}=\dfrac{11.25}{100}=0.1125\left(mol\right)\)
\(\Rightarrow n_C=0.1125\left(mol\right)\)
\(m_{bt}=m_{CO_2}+m_{H_2O}=7.65\left(g\right)\)
\(\Rightarrow n_{H_2O}=\dfrac{7.65-0.1125\cdot44}{18}=0.15\left(mol\right)\)
\(\Rightarrow n_H=0.15\cdot2=0.3\left(mol\right)\)
Ta có :
\(x=\dfrac{n_C}{n_X}=\dfrac{0.1125}{0.0375}=3\)
\(y=\dfrac{n_H}{n_X}=\dfrac{0.3}{0.0375}=8\)
\(M=12\cdot3+8+16z=60\left(g\text{/}mol\right)\)
\(\Rightarrow z=1\)
\(CT:C_3H_8O\)
a, \(C_2H_6O+3O_2\underrightarrow{t^o}2CO_2+3H_2O\)
\(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3+H_2O\)
Ta có: \(n_{CaCO_3}=\dfrac{14,4}{100}=0,144\left(mol\right)\)
Theo PT: \(n_{CO_2}=n_{CaCO_3}=0,144\left(mol\right)\Rightarrow m_{CO_2}=0,144.44=6,336\left(g\right)\)
b, \(n_{C_2H_6O}=\dfrac{1}{2}n_{CO_2}=0,072\left(mol\right)\)
\(\Rightarrow m_{C_2H_6O}=0,072.46=3,312\left(g\right)\)
\(\Rightarrow V_{C_2H_6O}=\dfrac{3,312}{0,8}=4,14\left(ml\right)\)
Độ rượu = \(\dfrac{4,14}{4,5}.100=92^o\)
Khi tính độ rượu thì mình cần phải ghi thêm % vô nữa hay sao bạn
$a\big)$
$n_{C_2H_5OH}=\frac{18,4}{46}=0,4(mol)$
$C_2H_5OH+3O_2\xrightarrow{t^o}2CO_2+3H_2O$
Theo PT: $n_{O_2}=3n_{C_2H_5OH}=1,2(mol)$
$\to V_{kk}=1,2.5.22,4=134,4(l)$
$b\big)$
$CO_2+Ca(OH)_2\to CaCO_3\downarrow+H_2O$
Theo PT: $n_{CaCO_3}=n_{CO_2}=2n_{C_2H_5OH}=0,8(mol)$
$\to m_{\downarrow}=m_{CaCO_3}=0,8.100=80(g)$
PTHH: \(C_2H_5OH+3O_2\xrightarrow[]{t^o}2CO_2+3H_2O_{ }\)
\(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3\downarrow+H_2O\)
a) Ta có: \(n_{CaCO_3}=\dfrac{160}{100}=1,6\left(mol\right)=n_{CO_2}\) \(\Rightarrow n_{O_2}=2,4\left(mol\right)\)
\(\Rightarrow V_{kk}=\dfrac{2,4\cdot22,4}{20\%}=268,8\left(l\right)\)
b) Theo PTHH: \(n_{C_2H_5OH}=\dfrac{1}{2}n_{CO_2}=0,8\left(mol\right)\)
\(\Rightarrow a=C\%_{C_2H_5OH}=\dfrac{0,8\cdot46}{50\cdot0,8}\cdot100\%=92\%=92^o\)
cho mik hỏi là khúc cuối ak công thức: C%C2H5OH =\(\dfrac{0,8.46}{50.0,8}\) . 100%, chỗ 50.0.8 là V.D là công thức gì vậy