Đáp án:

$a,V_{O_2}=10,304l.$

$b,V_{CO_2}=7,168l.$

Giải thích các bước giải:

$a,PTPƯ:$

$CH_4+2O_2\xrightarrow{t^o} CO_2+2H_2O$  $(1)$

$2C_2H_4+5O_2\xrightarrow{t^o} 4CO_2+4H_2O$  $(2)$

$V_{CH_4}=\dfrac{4,48.40}{100}=1,792l.$

$⇒V_{C_2H_4}=4,48-1,792=2,688l.$

$⇒n_{CH_4}=\dfrac{1,792}{22,4}=0,08mol.$

$⇒n_{C_2H_4}=\dfrac{2,688}{22,4}=0,12mol.$

$⇒n_{O_2}=2n_{CH_4}+\dfrac{5}{2}n_{C_2H_4}=0,08.2+0,12.\dfrac{5}{2}=0,46mol.$

$⇒V_{O_2}=0,46.22,4=10,304l.$

$b,n_{CO_2}=n_{CH_4}+2n_{C_2H_4}=0,08+0,12.2=0,32mol.$

$⇒V_{CO_2}=0,32.22,4=7,168l.$

chúc bạn học tốt!

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)

$n_{CO2}=nCH4=0,08(mol)$

$n_{CO2}=2n_{C2H2}=0,24(mol)$

=>$n_{CO2}=0,24+0,08=0,32(mol)$

=>$V_{CO2}=0,32.22,4=7,168(l)$