Đáp án:
Giải thích các bước giải:
$1/$
$PTPƯ:S+O_2\buildrel{{t^o}}\over\longrightarrow$ $SO_2$
$a,n_{S}=\frac{6,4}{32}=0,2mol.$
$n_{O_2}=\frac{9,6}{32}=0,3mol.$
$Vì$ $0,2<0,3$ $nên:$
$⇒O_2$ $dư.$
$⇒n_{O_2}(dư)=0,3-0,2=0,1mol.$
$⇒m_{O_2}(dư)=0,1.32=3,2g.$
$b,Theo$ $pt:$ $n_{SO_2}=n_{S}=0,2mol.$
$⇒V_{SO_2}=0,2.22,4=4,48l.$
$2/$
$PTPƯ:C+O_2\buildrel{{t^o}}\over\longrightarrow$ $CO_2$
$a,n_{C}=\frac{4,8}{12}=0,4mol.$
$n_{O_2}=\frac{44,8}{22,4}=2mol.$
$Vì$ $0,4<2$ $nên:$
$⇒O_2$ $dư.$
$⇒n_{O_2}(dư)=2-0,4=1,6mol.$
$⇒m_{O_2}(dư)=1,6.32=51,2g.$
$b,Theo$ $pt:$ $n_{CO_2}=n_{C}=0,4mol.$
$⇒V_{CO_2}=0,4.22,4=8,96l.$
$3/$
$PTPƯ:4P+5O_2\buildrel{{t^o}}\over\longrightarrow$ $2P_2O_5$
$a,n_{P}=\frac{12,4}{31}=0,4mol.$
$n_{O_2}=\frac{6,72}{22,4}=0,3mol.$
$Vì$ $0,4.5>0,3.4$ $nên:$
$⇒P$ $dư.$
$⇒n_{P}(dư)=0,4-(\frac{0,3.4}{5})=0,16mol.$
$⇒m_{P}(dư)=0,16.31=4,96g.$
$b,Theo$ $pt:$ $n_{P_2O_5}=\frac{2}{5}n_{O_2}=0,12mol.$
$⇒m_{P_2O_5}=0,12.142=17,04g.$
chúc bạn học tốt!