Đáp án:
Giải thích các bước giải:
$1/$
$PTPƯ:2H_2+O_2\buildrel{{t^o}}\over\longrightarrow$ $2H_2O$
$a,n_{H_2}=\frac{5,6}{22,4}=0,25mol.$
$Theo$ $pt:$ $n_{O_2}=\frac{1}{2}n_{H_2}=0,125mol.$
$⇒V_{O_2}=0,125.22,4=2,8l.$
$⇒m_{O_2}=0,125.32=4g.$
$b,Theo$ $pt:$ $n_{H_2O}=n_{H_2}=0,25mol.$
$⇒m_{H_2O}=0,25.18=4,5g.$
$2/$
$PTPƯ:Fe_2O_3+3H_2\buildrel{{t^o}}\over\longrightarrow$ $2Fe+3H_2O$
$n_{Fe_2O_3}=\frac{3,2}{160}=0,02mol.$
$Theo$ $pt:$ $n_{Fe}=2n_{Fe_2O_3}=0,04mol.$
$⇒m_{Fe}=0,04.56=2,24g.$
$b,Theo$ $pt:$ $n_{H_2}=3n_{Fe_2O_3}=0,06mol.$
$⇒V_{H_2}=0,06.22,4=1,344l.$
$3/$
$a,PbO+H_2\buildrel{{t^o}}\over\longrightarrow$ $Pb+H_2O$
$b,CuO+H_2\buildrel{{t^o}}\over\longrightarrow$ $Cu+H_2O$
$c,Fe_2O_3+3H_2\buildrel{{t^o}}\over\longrightarrow$ $2Fe+3H_2O$
$d,ZnO+H_2\buildrel{{t^o}}\over\longrightarrow$ $Zn+H_2O$
$e,Fe_3O_4+4H_2\buildrel{{t^o}}\over\longrightarrow$ $3Fe+4H_2O$
$f,MgO+H_2\buildrel{{t^o}}\over\longrightarrow$ $Mg+H_2O$
chúc bạn học tốt!