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

nCO₂ = nCaCO₃ = 0,9

Mà mCO₂ + mH₂O = 50,4  mH₂O = 50,4 – 0,9.44 = 10,8g  nH₂O = 0,6 mol

=>n_ankin = nCO₂ – nH₂O = 0,3 mol  V = 6,72 lít  

=>Chọn A.

a,PT: \(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)

Ta có: \(n_{CH_4}=\dfrac{3,2}{16}=0,2\left(mol\right)\)

Theo PT: \(n_{O_2}=2n_{CH_4}=0,4\left(mol\right)\)

\(\Rightarrow V_{O_2}0,4.22,4=8,96\left(l\right)\)

b, Theo PT: \(n_{CO_2}=n_{CH_4}=0,2\left(mol\right)\)

PT: \(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_{3\downarrow}+H_2O\)

Theo PT: \(n_{CaCO_3}=n_{CO_2}=0,2\left(mol\right)\)

\(\Rightarrow m_{\downarrow}=m_{CaCO_3}=0,2.100=20\left(g\right)\)

Bạn tham khảo nhé!

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

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) 

$SO_2 + Ca(OH)_2 \to CaSO_3 + H_2O$
$n_{SO_2} = n_{CaSO_3} = \dfrac{10}{120} = \dfrac{1}{12}(mol)$
$n_{N_2} = 0,3 - \dfrac{1}{12}=  \dfrac{13}{60}(mol)$
$\%m_{N_2} =$ \(\dfrac{\dfrac{13}{60}.28}{\dfrac{13}{60}.28+\dfrac{1}{12}.64}.100\%=53,22\%\)

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