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\(n_{CaCO_3}=\dfrac{100,2}{100}=1,002\left(mol\right)\)
PTHH: Ca(OH)2 + CO2 ---> CaCO3 + H2O
1,002 <---- 1,002
C2H5OH + 3O2 --to--> 2CO2 + 3H2O
0,501 <-------------------- 1,002
\(\rightarrow m_{C_2H_5OH}=0,501.46=23,046\left(g\right)\\ \rightarrow V_{C_2H_5OH}=\dfrac{23,046}{0,8}=28,8075\left(ml\right)\)
=> Độ rượu là: \(\dfrac{29,8075}{30}=96,025^o\)
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: \(n_{C_2H_5OH}=\dfrac{1}{2}n_{CO_2}=0,8\left(mol\right)\)
\(\Rightarrow a=C\%_{C_2H_5OH}=\dfrac{0,8\cdot46}{50\cdot0,8}\cdot100\%=92\%=92^o\)
\(n_{CaCO_3}=\dfrac{100}{100}=1mol\)
\(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3+H_2O\)
1 1 ( mol )
\(C_2H_5OH+3O_2\rightarrow\left(t^o\right)2CO_2+3H_2O\)
0,5 1 ( mol )
\(V_{C_2H_5OH}=\dfrac{0,5.46}{0,8}=28,75ml\)
Độ rượu = \(\dfrac{28,75}{30}.100=95,83độ\)
tham khảo
Theo PT , nC2H5OH = 1/2 nCO2 = 0,5 mol
⇒ mC2H5OH = 0,5.46 = 23 gam
⇒ V C2H5OH = 23/0,8 = 28,75 ml
⇒ Độ rượu : Đr = 28,75/30.100 = 96 độ
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)
$C_2H_5OH + 3O_2 \xrightarrow{t^o} 2CO_2 + 3H_2O$
$CO_2 + Ca(OH)_2 \xrightarrow{t^o} CaCO_3 + H_2O$
Theo PTHH :
$n_{CO_2} = n_{CaCO_3} = \dfrac{167}{100} = 1,67(mol)$
$n_{C_2H_5OH} = \dfrac{1}{2}n_{CO_2} = 0,835(mol)$
$m_{C_2H_5OH} = 0,835.46 = 38,41(gam)$
$V_{C_2H_5OH} = \dfrac{m}{D} = \dfrac{38,41}{0,8} = 48,0125(ml)$
Độ rượu $= \dfrac{48,0125}{60}.100 = 80,03^o$
b) $n_{O_2} = \dfrac{3}{2}n_{CO_2} = 2,505(mol)$
$V_{O_2} = 2,505.22,4 = 56,112(lít)$
$V_{kk} = 5V_{O_2} = 280,56(lít)$
C2H5OH + 3O2 => (to) 2CO2 + 3H2O
0.5 1.5 1 (mol)
CO2 + Ca(OH)2 => CaCO3 + H2O
1 1 (mol)
nCaCO3 = 100/100 = 1 (mol)
30 ml rượu = 0.03 l rượu
mC2H5OHpt=0.5.46 =23 (g)
V = m/D = 23/0.8 =28.75 (ml) =0.02875 (l);
Dr = 0.02875.100/0.03 = 95,8(3) xấp xỉ 96o
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 = \(\dfrac{28,75}{30}.100=95,83^o\)