Đáp án:

Giải thích các bước giải:

 a) $C2H4+Br2-->C2H4Br2$

$nBr2=0,56/160=0,0035(mol)$

$nC2H4=nBr2=0,0035(mol)$

=>$V_{C2H4}=0,003.22,4=0,0784(l)$

=>%$V_{C2H4}=0,0784/1,12.100$%$=7%$

=>%$VCH4=100-7=93$%

b)

$CH4+2O2-to->CO2+2H2O$

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

$nCO2=nCH4=0,0053(mol)$

$nCH4=1,12-0,0784/22,4=0,0465(mol)$

=>$nCO2=2nC2H4=0,093(mol)$

=>$nCO2=0,093+0,0035=0,0965(mol)$

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

$nCaCO3=nCO2=0,0965(mol)$

=>$mCaCO3=0,0965.100=9,65(g)$

Đáp án:

\(a,\ \%V_{C_2H_4}=7\%\\ \%V_{CH_4}=93\%\\ b,\ m_{CaCO_3}=5,35\ g.\)

Giải thích các bước giải:

\(a,\ PTHH:\ C_2H_4+Br_2 \xrightarrow{} C_2H_4Br_2\\ n_{Br_2}=\dfrac{0,56}{160}=0,0035\ mol.\\ Theo\ pt:\ n_{C_2H_4}=n_{Br_2}=0,0035\ mol.\\ ⇒\%V_{C_2H_4}=\dfrac{0,0035.22,4}{1,12}.100\%=7\%\\ ⇒\%V_{CH_4}=100\%-7\%=93\%\\ b,\ PTHH:\\ C_2H_4+3O_2 \xrightarrow{t^o} 2CO_2+2H_2O\ (1)\\ CH_4+2O_2 \xrightarrow{t^o} CO_2+2H_2O\ (2)\\ CO_2+Ca(OH)_2 \xrightarrow{} CaCO_3↓+H_2O\ (3)\\ n_{CH_4}=\dfrac{1,12}{22,4}-0,0035=0,0465\ mol.\\ Theo\ pt\ (1):\ n_{CO_2}=2n_{C_2H_4}=0,007\ mol.\\ Theo\ pt\ (2):\ n_{CO_2}=n_{CH_4}=0,0465\ mol.\\ ⇒n_{CO_2}(pt1+pt2)=0,007+0,0465=0,0535\ mol.\\ Theo\ pt\ (3):\ n_{CaCO_3}=n_{CO_2}=0,0535\ mol.\\ ⇒m_{CaCO_3}=0,0535.100=5,35\ g.\)

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