Đốt 115ml dd c2h5oh biết d=0,8g/ml tính a) v co2 sinh ra b) v o2 cần dùng 05/10/2021 Bởi Isabelle Đốt 115ml dd c2h5oh biết d=0,8g/ml tính a) v co2 sinh ra b) v o2 cần dùng
$a,PTPƯ:C_2H_5OH+3O_2\xrightarrow{t^o} 2CO_2+3H_2O$ $m_{C_2H_5OH}=115.0,8=92g.$ $⇒n_{C_2H_5OH}=\dfrac{92}{46}=2mol.$ $Theo$ $pt:$ $n_{CO_2}=2n_{C_2H_5OH}=4mol.$ $⇒V_{CO_2}=4.22,4=89,6l.$ $b,Theo$ $pt:$ $n_{O_2}=3n_{C_2H_5OH}=6mol.$ $⇒V_{O_2}=6.22,4=134,4l.$ chúc bạn học tốt! Bình luận
$m_{C_2H_5OH}=115.0,8=92g$ $⇒n_{C_2H_5OH}=92/46=2mol$ $PTHH :$ $C_2H_5OH+3O_2\overset{t^o}\to 2CO_2+3H_2O$ a.Theo pt : $n_{CO_2}=2.n_{C_2H_5OH}=2.2=4mol$ $⇒V_{CO_2}=4.22,4=89,6l$ b.Theo pt : $n_{O_2}=3.n_{C_2H_5OH}=3.2=6g$ $⇒V_{O_2}=6.22,4=134,4g$ Bình luận
$a,PTPƯ:C_2H_5OH+3O_2\xrightarrow{t^o} 2CO_2+3H_2O$
$m_{C_2H_5OH}=115.0,8=92g.$
$⇒n_{C_2H_5OH}=\dfrac{92}{46}=2mol.$
$Theo$ $pt:$ $n_{CO_2}=2n_{C_2H_5OH}=4mol.$
$⇒V_{CO_2}=4.22,4=89,6l.$
$b,Theo$ $pt:$ $n_{O_2}=3n_{C_2H_5OH}=6mol.$
$⇒V_{O_2}=6.22,4=134,4l.$
chúc bạn học tốt!
$m_{C_2H_5OH}=115.0,8=92g$
$⇒n_{C_2H_5OH}=92/46=2mol$
$PTHH :$
$C_2H_5OH+3O_2\overset{t^o}\to 2CO_2+3H_2O$
a.Theo pt :
$n_{CO_2}=2.n_{C_2H_5OH}=2.2=4mol$
$⇒V_{CO_2}=4.22,4=89,6l$
b.Theo pt :
$n_{O_2}=3.n_{C_2H_5OH}=3.2=6g$
$⇒V_{O_2}=6.22,4=134,4g$