Đáp án:
Giải thích các bước giải:
$1/$
$PTPƯ:Fe+CuSO_4→FeSO_4+Cu$
$n_{Fe}=\dfrac{11,2}{56}=0,2mol.$
$n_{CuSO_4}=\dfrac{40}{160}=0,25mol.$
$\text{Lập tỉ lệ:}$
$\dfrac{0,2}{1}<\dfrac{0,25}{1}$
$⇒n_{CuSO_4}$ $dư.$
$\text{⇒Tính theo}$ $n_{Fe}.$
$Theo$ $pt:$ $n_{Cu}=n_{Fe}=0,2mol.$
$⇒m_{Cu}=0,2.64=12,8g.$
$2/$
$a,PTPƯ:Fe+H_2SO_4→FeSO_4+H_2↑$
$n_{Fe}=\dfrac{22,4}{56}=0,4mol.$
$n_{H_2SO_4}=\dfrac{24,5}{98}=0,25mol.$
$\text{Lập tỉ lệ:}$
$\dfrac{0,4}{1}>\dfrac{0,25}{1}$
$⇒n_{Fe}$ $dư.$
$\text{⇒Tính theo}$ $n_{H_2SO_4}.$
$Theo$ $pt:$ $n_{H_2}=n_{H_2SO_4}=0,25mol.$
$⇒V_{H_2}=0,25.22,4=5,6l.$
$b,Theo$ $pt:$ $n_{FeSO_4}=n_{H_2SO_4}=0,25mol.$
$⇒m_{FeSO_4}=0,25.152=38g.$
$n_{Fe}(dư)=0,4-\dfrac{0,25.1}{1}=0,15mol.$
$⇒m_{Fe}(dư)=0,15.56=8,4g.$
$3/$
$a,PTPƯ:Zn+H_2SO_4→ZnSO_4+H_2↑$
$b,n_{Zn}=\dfrac{26}{65}=0,4mol.$
$n_{H_2SO_4}=\dfrac{49}{98}=0,5mol.$
$\text{Lập tỉ lệ:}$
$\dfrac{0,4}{1}<\dfrac{0,5}{1}$
$⇒n_{H_2SO_4}$ $dư.$
$\text{⇒Tính theo}$ $n_{Zn}.$
$Theo$ $pt:$ $n_{H_2}=n_{Zn}=0,4mol.$
$⇒V_{H_2}=0,4.22,4=8,96l.$
$c,Theo$ $pt:$ $n_{ZnSO_4}=n_{Zn}=0,4mol.$
$⇒m_{ZnSO_4}=0,4.161=64,4g.$
$n_{H_2SO_4}(dư)=0,5-\dfrac{0,4.1}{1}=0,1mol.$
$⇒m_{H_2SO_4}(dư)=0,1.98=9.8g.$
$4/$
$a,PTPƯ:CuO+2HCl→CuCl_2+H_2O$
$b,n_{CuO}=\dfrac{4}{80}=0,05mol.$
$n_{HCl}=\dfrac{2,92}{36,5}=0,08mol.$
$\text{Lập tỉ lệ:}$
$\dfrac{0,05}{1}<\dfrac{0,08}{1}$
$⇒n_{HCl}$ $dư.$
$\text{⇒Tính theo}$ $n_{CuO}.$
$Theo$ $pt:$ $n_{Cu}=n_{CuO}=0,05.64=3,2g.$
$Theo$ $pt:$ $n_{H_2O}=n_{CuO}=0,05mol.$
$⇒m_{H_2O}=0,05.18=0,9g.$
$n_{HCl}(dư)=0,08-\dfrac{0,05.1}{1}=0,03mol.$
$⇒m_{HCl}(dư)=0,03.36,5=1,095g.$
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