Đáp án:
$a,$
$m_{Fe}=16,8\ g\\m_{Cu}=3,2\ g.$
$\%m_{Fe}=84\%\\\%m_{Cu}=16\%$
$b,m_{CuO}=4\ g.$
Giải thích các bước giải:
$a,PTPƯ:Fe+2HCl\xrightarrow{} FeCl_2+H_2↑$
$n_{H_2}=\dfrac{6,72}{22,4}=0,3\ mol.$
$Theo$ $pt:$ $n_{Fe}=n_{H_2}=0,3\ mol.$
$⇒m_{Fe}=0,3.56=16,8\ g.$
$⇒m_{Cu}=20-16,8=3,2\ g.$
$⇒\%m_{Fe}=\dfrac{16,8}{20}.100\%=84\%$
$⇒\%m_{Cu}=\dfrac{3,2}{20}.100\%=16\%$
$b,PTPƯ:CuO+CO\xrightarrow{t^o} Cu+CO_2↑$
$n_{Cu}=\dfrac{3,2}{64}=0,05\ mol.$
$Theo$ $pt:$ $n_{CuO}=n_{Cu}=0,05\ mol.$
$⇒m_{CuO}=0,05.80=4\ g.$
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