Đáp án:
Giải thích các bước giải:
a) $CH4+2O2-to->CO2+2H2O$
$2C2H2+5O2-to->4CO2+2H2O$
$V_{CH4}=4,48.40$%$=1,792(l)$
=>$n_{CH4}=1,792/22,4=0,08(mol)$
$n_{O2}=2n_{CH4}=0,16(mol)$
$n_{C2H2}=4,48-1,792/22,4=0,12(mol)$
$n_{O2}=5/2n_{C2H2}=0,3(mol)$
=>$n_{O2}=0,3+0,16=0,46(mol)$
=>$V_{O2}=0,46.22,4=10,304(l)$
b)
$n_{CO2}=nCH4=0,08(mol)$
$n_{CO2}=2n_{C2H2}=0,24(mol)$
=>$n_{CO2}=0,24+0,08=0,32(mol)$
=>$V_{CO2}=0,32.22,4=7,168(l)$