CH4+2O2−−>CO2+2H2OCH4+2O2−−>CO2+2H2O
x-------------------------x(mol)
C2H2+52O2−−>2CO2+H2OC2H2+52O2−−>2CO2+H2O
y-------------------------------2y(mol)
CO2+Ca(OH)2−−>CaCO3+H2OCO2+Ca(OH)2−−>CaCO3+H2O
nCaCO3=60100=0,6(mol)nCaCO3=60100=0,6(mol)
nCO2=nCaCO3=0,6(mol)nCO2=nCaCO3=0,6(mol)
nhh=11.2224=0,5(mol)nhh=11.2224=0,5(mol)
Theo bài ta có hpt
{x+y=0,5x+2y=0,6⇒{x=0,4y=0,1{x+y=0,5x+2y=0,6⇒{x=0,4y=0,1
%nCH4=0,40,5.100%=80%%nCH4=0,40,5.100%=80%
%nC2H2=100−80=20%