\(PTHH:C_2H_5OH+3O_2-^{t^o}\rightarrow2CO_2+3H_2O\)
\(n_{C_2H_5OH}=\frac{13,8}{46}=0,3\left(mol\right)\)
Theo PTHH: \(n_{O_2}=3n_{C_2H_5OH}=0,9\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,9.22,4=20,16\left(l\right)\)
\(\Rightarrow V_{kk}=\frac{20,16.100}{20}=100,8\left(l\right)\)
b)Theo PT: \(n_{CO_2}=2n_{C_2H_5OH}=0,6\left(mol\right)\)
\(n_{H_2O}=3n_{C_2H_5OH}=0,9\left(mol\right)\)
\(m_{CO_2}=0,6.44=26,4\left(g\right)\)
\(m_{H_2O}=0,9.18=16,2\left(g\right)\)
c)\(n_{Ca\left(OH\right)_2}=\frac{0,45}{1}=0,45\left(mol\right)\)
Ta có \(\frac{n_{Ca\left(OH\right)_2}}{n_{CO_2}}=\frac{0,45}{0,6}=0,75\)
\(\Rightarrow\)Chỉ tạo ra muối CaCO3
\(PTHH:CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3+H_2O\)
Trước.p/ứ:0,6.....0,45
P.ứ.........0,45......0,45
Sau p.ứ:...0,15......0
=>CO2 dư, Ca(OH)2 p.ứ hết
Theo PT:\(n_{CaCO_3}=n_{Ca\left(OH\right)_2}=0,45\left(mol\right)\)
\(\Rightarrow m_{CaCO_3}=0,45.100=45\left(g\right)\)
C2H5OH + \(\frac{7}{2}\)O2 \(\underrightarrow{to}\) 2CO2 + 3H2O (1)
\(n_{C_2H_5OH}=\frac{13,8}{46}=0,3\left(mol\right)\)
a) Theo Pt1: \(n_{O_2}=\frac{7}{2}n_{C_2H_5OH}=\frac{7}{2}\times0,3=1,05\left(mol\right)\)
\(\Rightarrow V_{O_2}=1,05\times22,4=23,52\left(l\right)\)
\(\Rightarrow V_{KK}=\frac{23,52}{20\%}=117,6\left(l\right)\)
b) \(m_{O_2}=1,05\times32=33,6\left(g\right)\)
Theo ĐL BTKL ta có: \(m_{C_2H_5OH}+m_{O_2}=m_{CO_2}+m_{H_2O}\)
\(\Leftrightarrow13,8+33,6=m_{sp}\)
\(\Leftrightarrow m_{sp}=47,4\left(g\right)\)
c) CO2 + Ca(OH)2 → CaCO3↓ + H2O (2)
\(n_{Ca\left(OH\right)_2}=0,45\times1=0,45\left(mol\right)\)
Theo Pt1: \(n_{CO_2}=2n_{C_2H_5OH}=2\times0,3=0,6\left(mol\right)\)
Theo Pt2: \(n_{CO_2}=n_{Ca\left(OH\right)_2}\)
Theo bài: \(n_{CO_2}=\frac{4}{3}n_{Ca\left(OH\right)_2}\)
Vì \(\frac{4}{3}>1\) ⇒ CO2 dư ⇒ pư tiếp
CO2 + CaCO3 + H2O → Ca(HCO3)2 (3)
Theo PT2: \(n_{CO_2}pứ=n_{Ca\left(OH\right)_2}=0,45\left(mol\right)\)
\(\Rightarrow n_{CO_2}dư=n_{CO_2}\left(3\right)=0,6-0,45=0,15\left(mol\right)\)
Theo Pt2: \(n_{CaCO_3}=n_{Ca\left(OH\right)_2}=0,45\left(mol\right)\)
Theo pT3: \(n_{CO_2}=n_{CaCO_3}\)
Theo bài: \(n_{CO_2}=\frac{1}{3}n_{CaCO_3}\)
Vì \(\frac{1}{3}< 1\) ⇒ CaCO3 dư
Theo pT3: \(n_{CaCO_3}pư=n_{CO_2}=0,15\left(mol\right)\)
\(\Rightarrow n_{CaCO_3}dư=0,45-0,15=0,3\left(mol\right)\)
\(\Rightarrow m_{CaCO_3}dư=0,3\times100=30\left(g\right)\)
Theo Pt3: \(n_{Ca\left(HCO_3\right)_2}=n_{CO_2}=0,15\left(mol\right)\)
\(\Rightarrow m_{Ca\left(HCO_3\right)_2}=0,15\times162=24,3\left(g\right)\)
Vậy \(m_{muối}=24,3+30=54,3\left(g\right)\)
nC2H5OH = 13.8/46=0.3 mol
C2H5OH + 3O2 -to-> 2CO2 + 3H2O
0.3________0.9______0.6_____0.9
Vkk= 5VO2 = 5*0.9*22.4 = 100.8 l
mCO2 = 0.6*44=26.4 g
mH2O = 0.9*18= 16.2 g
nCa(OH)2 = 0.45 mol
nCO2/nCa(OH)2 = 0.6/0.45=1.333
=> Tạo ra 2 muối
Đặt :
nCaCO3 = x mol
nCa(HCO3)2 = y mol
Ca(OH)2 + CO2 --> CaCO3 + H2O
x__________x________x
Ca(OH)2 + 2CO2 --> Ca(HCO3)2
y__________2y________y
<=> x + y = 0.45
x + 2y = 0.6
=> x = 0.3
y = 0.15
mCaCO3 = 30g
mCa(HCO3)2 = 24.3 g