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Trả lời:
Bài 15:
$TH2:$
$Pt:\,Fe+2HCl \rightarrow FeCl_2+H_2$
$n_{FeCl_2}=\dfrac{25,4}{127}=0,2\,(mol)$
$⇒n_{Fe}=0,2\,(mol)$
$TH1:$
$Pt_1:\,2Fe+3Cl_2 \rightarrow 2FeCl_3\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,2$
$Pt_2:\,Cu+Cl_2 \rightarrow CuCl_2$
$m_{FeCl_3}=0,2.162,5=32,5\,(g)$
$⇒m_{CuCl_2}=59,5-32,5=27\,(g)$
$⇒n_{CuCl_2}=0,2\,(mol)$
$⇒n_{Cu}=0,2\,(mol)$
$m_{X}=64.0,2+56.0,2=24\,(g).$
Bài 16:
$Pt_1:\,Mg+2HCl \rightarrow MgCl_2+H_2$
$Pt_2:\,Fe+2HCl \rightarrow FeCl_2+H_2$
$n_{H_2}=\dfrac{1}{2}=0,5\,(mol)$
$⇒n_{HCl}=2.n_{H_2}=1\,(mol)$
Áp dụng ĐLBTKL:
$⇒m_{muối}=m_{h^2}+m_{HCl}-m{H_2}=20+1.36,5-0,5.2=55,5\,(g).$
