$a,PTPƯ:C_2H_5OH+3O_2\xrightarrow{t^o} 2CO_2+3H_2O$
$n_{C_2H_5OH}=\dfrac{4,6}{46}=0,1mol.$
$Theo$ $pt:$ $n_{CO_2}=2n_{C_2H_5OH}=0,2mol.$
$⇒m_{CO_2}=0,2.44=8,8g.$
$b,Theo$ $pt:$ $n_{O_2}=3n_{C_2H_5OH}=0,3mol.$
$⇒V_{O_2}=0,3.22,4=6,72l.$
$c,V_{CO_2}=0,2.22,4=4,48l.$
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