1)

 a)

Fe+2HCl->FeCl2+H2

b)

nH2=4,48/22,4=0,2 mol

=>nFe=0,2 mol

%mFe=0,2x56/30,4x100%=36,84%

%mCu=100-36,84=63,16%

c)

CMHCl=0,4/0,5=0,8 M

2)

a)

Mg+2HCl->MgCl2+H2

b)

mCu=1,28 g

%mCu=1,28/20,488x100%=6,25%

%mMg=100-6,25=93,75%

c)

nMg=0,8 mol=>nH2=0,8 mol

VH2=0,8x22,4=17,92 l

CMHCl=1,6/0,4=4 M

$\text{Câu 1 :}$

$n_{H_2}=\dfrac{4,48}{22,4}=0,2mol$

$a.Fe+2HCl\to FeCl_2+H_2↑$

$\text{Cu không phản ứng với HCl}$

$\text{b.Theo pt :}$

$n_{Fe}=n_{H_2}=0,2mol$

$⇒m_{Fe}=0,2.56=11,2g$

$\%m_{Fe}=\dfrac{11,2}{30,4}.100\%=36,84\%$

$\%m_{Cu}=100\%-36,84\%=63,16\%$

$n_{HCl}=2.n_{H_2}=0,2.2=0,4mol$

$⇒C_{M_{HCl}}=\dfrac{0,4}{0,5}=0,8M$

$\text{Câu 2 :}$

$a.Mg+2HCl\to MgCl_2+H_2↑$

$\text{Cu không phản ứng với HCl}$

$b.m_{rắn}=m_{Cu}=1,28g$

$⇒\%m_{Cu}\dfrac{1,28}{20,488}.100\%=6,25\%$

$\%m_{Mg}=100\%-6,25\%=93,75\%$

$c.m_{Mg}=20,488-1,28=19,208g$

$⇒n_{Mg}=\dfrac{19,208}{24}≈0,8mol$

$\text{Theo pt :}$

$n_{H_2}=n_{Mg}=0,8mol$

$⇒V=V_{H_2}=0,8.22,4=17,92l$

$n_{HCl}=2.n_{Mg}=2.0,8=1,6mol$

$⇒C_{M_{HCl}}=\dfrac{1,6}{0,4}=4M$