\(n_{CH_4}=\dfrac{m_{CH_4}}{M_{CH_4}}=\dfrac{1,6}{16}=0,1mol\)

\(CH_4+2O_2\rightarrow\left(t^o\right)CO_2+2H_2O\)

 0,1       0,2              0,1                   ( mol )

\(Ca\left(OH\right)_2+CO_2\rightarrow CaCO_3+H_2O\)

                     0,1         0,1                   ( mol )

\(m_{CaCO_3}=n_{CaCO_3}.M_{CaCO_3}=0,1.100=10g\)

\(V_{O_2}=n_{O_2}.22,4=0,2.22,4=4,48l\)

\(V_{kk}=V_{O_2}.5=4,48.5=22,4l\)

nCH4 = 1,6/16 = 0,1 (mol)

PTHH: CH4 + 2O2 -> (t°) CO2 + 2H2O

Mol: 0,1 ---> 0,2 ---> 0,1

CO2 + Ca(OH)2 -> CaCO3 + H2O

Mol: 0,1 ---> 0,1 ---> 0,1

mCaCO3 = 0,1 . 100 = 10 (g)

Vkk = 0,2 . 5 . 22,4 = 22,4 (l)

\(C_4H_{10}+\dfrac{13}{2}O_2-t^0->4CO_2+5H_2O\\ a.V_{KK}=\dfrac{5,6\cdot\dfrac{13}{2}}{0,2}=182\left(L\right)\\ b.CO_2+Ca\left(OH\right)_2->CaCO_3+H_2O\\ m_{KT}=100\cdot\dfrac{5,6}{22,4}\cdot4=100g\)

\(n_{C_2H_4}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\\ C_2H_4+3O_2\rightarrow\left(t^o\right)2CO_2+2H_2O\\ n_{CO_2}=2.n_{C_2H_4}=2.0,5=1\left(mol\right)\\ CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3+H_2O\\ n_{CaCO_3}=n_{CO_2}=1\left(mol\right)\\ m_{kết.tủa}=m_{CaCO_3}=100.1=100\left(g\right)\\ n_{O_2}=3.n_{C_2H_4}=3.0,5=1,5\left(mol\right)\\ V_{kk}=\dfrac{100}{20}.V_{O_2\left(đktc\right)}=5.\left(1,5.22,4\right)=168\left(lít\right)\)

nC2H4 = 8.4/0.3 (mol) 

C2H4 + 3O2 -to-> 2CO2 + 2H2O 

0.3_____0.9______0.6 

VO2 = 0.9*22.4 = 20.16 (l) 

Ca(OH)2 + CO2 => CaCO3 + H2O 

__________0.6______0.6 

mCaCO3 = 0.6*100 = 60 (g) 

C2H4+3O2-to>2CO2+2H2O

0,25----0,75------0,5

CO2+Ca(OH)2->CaCO3+H2O

0,5-----------------------0,5

n C2H4=0,25 mol

=>Vkk=0,75.22,4.5=84l

b)m CaCO3=0,5.100=50g

a) \(n_{hh}=\dfrac{4,48}{22,4}=0,2\left(mol\right);n_{CaCO_3}=\dfrac{56}{100}=0,56\left(mol\right)\)

Gọi \(\left\{{}\begin{matrix}n_{C_2H_6}=a\left(mol\right)\\n_{C_4H_{10}}=b\left(mol\right)\end{matrix}\right.\Rightarrow a+b=0,2\left(1\right)\)

PTHH: 

\(2C_2H_6+7O_2\xrightarrow[]{t^o}4CO_2+6H_2O\)

a------------------->2a

\(2C_4H_{10}+13O_2\xrightarrow[]{t^o}8CO_2+10H_2O\)

b---------------------->4b

\(Ca\left(OH\right)_2+CO_2\rightarrow CaCO_3\downarrow+H_2O\)

                     0,56<---0,56

\(\Rightarrow2a+4b=0,56\left(2\right)\)

Từ (1), (2) => a = 0,12; b = 0,08

\(\Rightarrow\left\{{}\begin{matrix}\%V_{C_2H_6}=\%n_{C_2H_6}=\dfrac{0,12}{0,2}.100\%=60\%\\\%V_{C_4H_{10}}=100\%-60\%=40\%\end{matrix}\right.\)

Và \(\left\{{}\begin{matrix}\%m_{C_2H_6}=\dfrac{0,12.30}{0,12.30+0,08.58}.100\%=43,67\%\\\%m_{C_4H_{10}}=100\%-43,67\%=56,33\%\end{matrix}\right.\)

Quy đổi C₄H₄ thành C₂H₂

\(n_{CaCO_3}=\dfrac{20}{100}=0,2\left(mol\right)\)

PTHH: 2C₂H₂ + 5O₂ --t^o--> 4CO₂ + 2H₂O

               0,1<--0,25<--------0,2

            Ca(OH)₂ + CO₂ --> CaCO₃ + H₂O

                              0,2<------0,2

=> \(m_X=0,1.26=2,6\left(g\right)\)

\(V_{O_2}=0,25.22,4=5,6\left(l\right)\)

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) $n_{O_2} = \dfrac{3}{2}n_{CO_2} = 2,505(mol)$
$V_{O_2} = 2,505.22,4 = 56,112(lít)$
$V_{kk} = 5V_{O_2} = 280,56(lít)$