Đáp án:
Giải thích các bước giải:
$3/$
$a,PTPƯ:Fe_2O_3+3H_2\buildrel{{t^o}}\over\longrightarrow$ $2Fe+3H_2O$
$b,n_{Fe_2O_3}=\dfrac{40}{160}=0,25mol.$
$Theo$ $pt:$ $n_{H_2}=3n_{Fe_2O_3}=0,75mol.$
$⇒V_{H_2}=0,25.22,4=16,8l.$
$c,Theo$ $pt:$ $n_{Fe}=2n_{Fe_2O_3}=0,5mol.$
$⇒m_{Fe}=0,5.56=28.$
$4/$
$PTPƯ:2Al+6HCl→2AlCl_3+3H_2↑$
$n_{Al}=\dfrac{10,8}{27}=0,6mol.$
$n_{HCl}=\dfrac{14,6}{36,5}=0,4mol.$
$\text{Lập tỉ lệ:}$
$\dfrac{0,6}{2}>\dfrac{0,4}{6}$
$⇒n_{Al}$ $dư.$
$⇒n_{Al}(dư)=0,6-\dfrac{0,4.2}{6}=\dfrac{7}{15}mol.$
$⇒m_{Al}(dư)=\dfrac{7}{15}.27=12,6g.$
$Theo$ $pt:$ $n_{H_2}=\dfrac{1}{2}n_{HCl}=0,2mol.$
$⇒V_{H_2}=0,2.22,4=4,48l.$
$5/$
$PTPƯ:4P+5O_2\buildrel{{t^o}}\over\longrightarrow$ $2P_2O_5$
$n_{P}=\dfrac{6,2}{31}=0,2mol.$
$n_{O_2}=\dfrac{6,4}{32}=0,2mol.$
$\text{Lập tỉ lệ:}$
$\dfrac{0,2}{4}<\dfrac{0,2}{5}$
$⇒n_{P}$ $dư.$
$⇒n_{P}(dư)=0,2-\dfrac{0,2.4}{5}=0,04mol.$
$⇒m_{P}(dư)=0,04.31=1,24g.$
$Theo$ $pt:$ $n_{P_2O_5}=\dfrac{2}{5}n_{O_2}=0,08mol.$
$⇒m_{P_2O_5}=0,08.142=11,36g.$
$6/$
$PTPƯ:CuO+H_2\buildrel{{t^o}}\over\longrightarrow$ $H_2O+Cu$
$n_{CuO}=\dfrac{40}{80}=0,5mol.$
$n_{H_2}=\dfrac{8,96}{22,4}=0,4mol.$
$\text{Lập tỉ lệ:}$
$\dfrac{0,5}{1}>\dfrac{0,4}{1}$
$⇒n_{CuO}$ $dư.$
$\text{⇒Tính theo}$ $n_{H_2}$
$Theo$ $pt:$ $n_{Cu}=n_{H_2}=0,4mol.$
$⇒m_{Cu}=0,4.64=25,6g.$
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