\(a.n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ n_{HCl}=0,2.1,5=0,3\left(mol\right)\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ Vì:\dfrac{0,3}{2}< \dfrac{0,2}{1}\\ \Rightarrow Mgdư\\ n_{H_2}=n_{MgCl_2}=\dfrac{0,3}{2}=0,15\left(mol\right)\\ V_{H_2\left(đktc\right)}=0,15.22,4=3,36\left(l\right)\\ b.V_{ddsau}=V_{ddHCl}=0,2\left(l\right)\\ C_{MddMgCl_2}=\dfrac{0,15}{0,2}=0,75\left(M\right)\)

a)

Gọi $n_{Fe} = a(mol) ; n_{Al} =b (mol) \Rightarrow 56a + 27b = 11(1)$
$Fe + 2HCl \to FeCl_2 + H_2$
$2Al + 6HCl \to 2AlCl_3 + 3H_2$
Theo PTHH : $n_{H_2} = a + 1,5b = \dfrac{8,96}{22,4} = 0,4(2)$

Từ (1)(2) suy ra : a = 0,1 ; b = 0,2

$\%m_{Fe} = \dfrac{0,1.56}{11}.100\% = 50,9\%$
$\%m_{Al} = 100\% - 50,9\% = 49,1\%$

b) $n_{HCl} = 2n_{H_2} = 0,8(mol)$
$\Rightarrow C_{M_{HCl}} = \dfrac{0,8}{0,4} = 2M$

c)

$C_{M_{FeCl_2}} = \dfrac{0,1}{0,4} = 0,25M$
$C_{M_{AlCl_3}} =\dfrac{0,2}{0,4} = 0,5M$

n_Al = 5.4/27 = 0.2  (mol) 

2Al + 3H2SO4 => Al2(SO4)3 + 3H2 

0.2____0.3_________________0.3 

VH2 = 0.3*22.4 = 6.72(l) 

CM H2SO4 = 0.3/0.1 = 3 M

PTHH: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\uparrow\)

Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\) 

\(\Rightarrow n_{H_2}=0,3mol=n_{H_2SO_4}\) \(\Rightarrow\left\{{}\begin{matrix}V_{H_2}=0,3\cdot22,4=6,72\left(l\right)\\C_{M_{H_2SO_4}}=\dfrac{0,3}{0,1}=3\left(M\right)\end{matrix}\right.\)

\(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\)

100ml = 0,1l

\(n_{HCl}=3.0,1=0,3\left(mol\right)\)

Pt : \(Mg+2HCl\rightarrow MgCl_2+H_2|\)

        1          2              1          1

      0,1        0,3           0,1        0,1

a) Lập tỉ số so sánh : \(\dfrac{0,1}{1}< \dfrac{0,3}{2}\)

               ⇒ Mg phản ứng hết , HCl dư

               ⇒ Tính toán dựa vào số mol của Mg

\(n_{H2}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)

\(V_{H2\left(dktc\right)}=0,1.22,4=2,24\left(l\right)\)

b) \(n_{MgCl2}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)

\(n_{HCl\left(dư\right)}=0,3-\left(0,1.2\right)=0,1\left(mol\right)\)

\(C_{M_{MgCl2}}=\dfrac{0,1}{0,1}=1\left(M\right)\)

\(C_{M_{HCl\left(dư\right)}}=\dfrac{0,1}{0,1}=1\left(M\right)\)

 Chúc bạn học tốt

\(n_{Fe}=\dfrac{84}{56}=1,5\left(mol\right)\\ PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\\ n_{H_2}=n_{FeCl_2}=n_{Fe}=1,5\left(mol\right)\\ V_{H_2}=1,5.22,4=33,6\left(l\right)\\ C\%_{ddFeCl_2}=\dfrac{127.1,5}{84+300-1,5.2}.100\%=\dfrac{190,5}{381}.100\%=50\%\)

\(n_{Zn}=\dfrac{130}{65}=2mol\\ Zn+2HCl\rightarrow ZnCl_2+H_2\\ n_{Zn}=n_{H_2}=2mol\\ V_{H_2}=2.22,4=44,8l\\ 1000ml=1l\\ n_{HCl}=2.2=4mol\\ C_{M_{HCl}}=\dfrac{4}{1}=4M\)

\(n_{Zn}=\dfrac{m_{Zn}}{M_{Zn}}=\dfrac{130}{65}=2mol\) 

PTHH: Zn + 2HCl \(\rightarrow\) ZnCl₂ + H₂

TL;       1        2          1          1

mol:      2 \(\rightarrow\) 4

\(m_{HCl}=n.M=4.36,5=146g\) 

đổi 1000 ml= 1l

\(C\%_{ddHCl}=\dfrac{m_{HCl}}{V_{HCl}}.100\%=\dfrac{146}{1}.100=14600\%\)

số hơi lớn em xem lại đề nhé

a) 2Al + 6HCl -> 2AlCl3 + 3H2

b)nAl = 2.7/27=0.1mol

2Al + 6HCl -> 2AlCl3 + 3H2

(mol) 0.1 0.3 0.1 0.15

VH2 = 0.15*22.4=3.36l

mAlCl3= 0.1*133.5=13.35g

=>C%=\(\dfrac{13,35}{500+27}100=2,533\%\)