Đáp án:
Giải thích các bước giải:
$3/$
$1.PTPƯ:2H_2+O_2\buildrel{{t^o}}\over\longrightarrow$ $2H_2O$
$n_{H_2}=\frac{22,4}{22,4}=1mol.$
$Theo$ $pt:$ $n_{O_2}=\frac{1}{2}n_{H_2}=0,5mol.$
$⇒V_{O_2}=0,5.22,4=11,2l.$
$2.PTPƯ:Zn+2HCl→ZnCl_2+H_2↑$
$Theo$ $pt:$ $n_{H_2}=n_{Zn}=0,2mol.$
$⇒V_{H_2}=0,2.22,4=4,48l.$
$3.PTPƯ:2H_2+O_2\buildrel{{t^o}}\over\longrightarrow$ $2H_2O$
$\text{Theo pt ta thấy nH2=2nO2⇒ VH2=2VO2}$
$\text{⇒Khí H2 còn dư sau phản ứng.}$
$4.PTPƯ:Fe+H_2SO_4→FeSO_4+H_2↑$
$n_{Fe}=\frac{11,2}{56}=0,2mol.$
$Theo$ $pt:$ $n_{H_2}=n_{Fe}=0,2mol.$
$⇒V_{H_2}=0,2.22,4=4,48l.$
$5.PTPƯ:4Na+O_2\buildrel{{t^o}}\over\longrightarrow$ $2Na_2O$
$n_{O_2}=\frac{8,96}{22,4}=0,4mol.$
$Theo$ $pt:$ $n_{Na}=4n_{O_2}=1,6mol.$
$⇒m_{Na}=1,6.23=36,8g.$
$4/$
$1.$
$-PTPƯ:Zn+2HCl→ZnCl_2+H_2↑$
$-n_{Zn}=\frac{39}{65}=0,6mol.$
$Theo$ $pt:$ $n_{H_2}=n_{Zn}=0,6mol.$
$⇒V_{H_2}=0,6.22,4=13,44l.$
$Theo$ $pt:$ $n_{ZnCl_2}=n_{Zn}=0,6mol.$
$⇒m_{ZnCl_2}=0,6.136=81,6g.$
$2.$
$-PTPƯ:Fe+2HCl→FeCl_2+H_2↑$
$-n_{Fe}=\frac{33,6}{56}=0,6mol.$
$Theo$ $pt:$ $n_{H_2}=n_{Fe}=0,6mol.$
$⇒V_{H_2}=0,6.22,4=13,44l.$
$Theo$ $pt:$ $n_{FeCl_2}=n_{Fe}=0,6mol.$
$⇒m_{FeCl_2}=0,6.127=76,2g.$
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