a)
$C_2H_5OH + 3O_2 \xrightarrow{t^o} 2CO_2 + 3H_2O$
$CO_2 + Ca(OH)_2 \to CaCO_3 + H_2O$
b)
$n_{CO_2} = n_{CaCO_3} = \dfrac{150}{100} = 1,5(mol)$
$n_{O_2} = \dfrac{3}{2}n_{O_2} = 2,25(mol)$
$V_{O_2} = 2,25.22,4 = 50,4(lít)$
$V_{không\ khí} = V_{O_2} : 20\% = 50,4:20\% = 252(lít)$
c) $n_{C_2H_5OH} = dfrac{1}{2}n_{CO_2} = 0,75(mol)$
$m_{C_2H_5OH} = 0,75.46 = 34,5(gam)$
$V_{C_2H_5OH} = \dfrac{m}{D} = \dfrac{34,5}{0,8} = 43,125(ml)$
Độ rượu : $Đ_{C_2H_5OH} = \dfrac{43,125}{45}.100 = 95,83$
a, \(C_2H_6O+3O_2\underrightarrow{t^o}2CO_2+3H_2O\)
\(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3+H_2O\)
b, Ta có: \(n_{CaCO_3}=\dfrac{150}{100}=1,5\left(mol\right)\)
Theo PT: \(n_{O_2}=\dfrac{3}{2}n_{CO_2}=\dfrac{3}{2}n_{CaCO_3}=2,25\left(mol\right)\Rightarrow V_{O_2}=2,25.22,4=50,4\left(l\right)\)
\(\Rightarrow V_{kk}=\dfrac{V_{O_2}}{20\%}=252\left(l\right)\)
c, Theo PT: \(n_{C_2H_6O}=\dfrac{1}{2}n_{CO_2}=\dfrac{1}{2}n_{CaCO_3}=0,75\left(mol\right)\)
\(\Rightarrow m_{C_2H_6O}=0,75.46=34,5\left(g\right)\)
\(\Rightarrow V_{C_2H_6O}=\dfrac{34,5}{0,8}=43,125\left(ml\right)\)
⇒ Độ rượu = \(\dfrac{43,125}{45}.100\approx95,833^o\)
