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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