a,\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
PTHH: Fe + 2HCl → FeCl₂ + H₂

Mol:     0,1    0,2                             0,1

b,\(m_{Fe}=0,1.56=5,6\left(g\right)\)

\(\Rightarrow\%m_{Fe}=\dfrac{5,6.100\%}{12}=46,67\%;\%m_{Cu}=100-46,67=53,33\%\)

c,\(m_{HCl}=0,2.36,5=7,3\left(g\right)\Rightarrow m_{ddHCl}=\dfrac{7,3.100}{14,6}=50\left(g\right)\)

a) Mg + 2HCl -> MgCl2 + H2
Al + 3HCl -> AlCl3 + 3/2H2

b) Gọi a, b lần lượt là số mol Mg, Al.

nH2 = 5,6/22,4 = 0,25 (mol)

Mg + 2HCl -> MgCl2 + H2

a        2a            a          a
Al + 3HCl -> AlCl3 + 3/2H2

b           3b       b          3/2b

Ta có hệ pt: 

mhh = 24a + 27b = 5,1 (g)

nH2 = a + 3/2b = 0,25 (mol)

=> a = 0,1 (mol)

b = 0,1 (mol)

200 ml = 0,2 l

nHCl = 2a + 3b = 0,2 + 0,3 = 0,5 (mol)

=> CM ddHCl = 0,5/0,2 = 2,5 (M)

%mMg = 24a/5,1*100% = 2,4/5,1*100% = 47,06%

%mAl = 100%-47,06% = 52,94%

a. PTHH : Mg + 2HCl ➝ MgCl₂ + H₂ (1)

b. theo bài : nH₂ = 3,36 : 22,4 = 0,15 (mol)

theo (1) nMg = nH₂ = 0,15 (mol)

➞ mMg = 0,15 ✖ 24 = 3,6 (g)

➞ %mMg = (3,6 : 5)✖100 = 72%

➞ %mCu = 100% - 72% = 28%

c. theo (1) nHCl = 2nH₂ = 2✖0,15 = 0,3 (mol)

mHCl = 0,3✖36,5 = 10,95(g)

➜mddHCl = (10,95✖100):14,6 = 75(g)

d. dung dịch Y : MgCl₂

mdd_(spư)= 3,6+75-0,3 = 78,3(g)

theo (1) nMgCl₂ = nH₂ = 0,15(mol)

mMgCl₂ = 0,15✖95 = 14,25(g)

C%MgCl₂ = (14,25 : 78,3)✖100 = 18,199%

Gọi nFe = a (mol); nMg = b (mol)

56a + 24b = 8 (1)

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

PTHH: 

Fe + 2HCl -> FeCl2 + H2

a ---> a ---> a ---> a

Mg + 2HCl -> MgCl2 + H2

b ---> b ---> b ---> b

a + b = 0,2 (2)

(1)(2) => a = b = 0,1 (mol)

mFe = 0,1 . 56 = 5,6 (g)

%mFe = 5,6/8 = 70%

%mMg = 100% - 70% = 30%

nHCl = 0,1 . 2 + 0,1 . 2 = 0,4 (mol)

CMddHCl = 0,4/0,1 = 4M

\(n_{H_2}=\dfrac{6,72}{22,4}=0,3mol\)

\(\left\{{}\begin{matrix}n_{Mg}=x\left(mol\right)\\n_{Zn}=y\left(mol\right)\end{matrix}\right.\Rightarrow24x+65y=11,3\left(1\right)\)

\(Mg+2HCl\rightarrow MgCl_2+H_2\)

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

\(\Rightarrow x+y=0,3\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,2mol\\y=0,1mol\end{matrix}\right.\)

a)\(\%m_{Mg}=\dfrac{0,2\cdot24}{11,3}\cdot100\%=42,48\%\)

\(\%m_{Zn}=100\%-42,48\%=57,52\%\)

b)\(n_{HCl}=2\left(n_{Mg}+n_{Zn}\right)=2\cdot\left(0,2+0,1\right)=0,6mol\)

\(C_{M_{HCl}}=\dfrac{0,6}{0,2}=3M\)

a.\(m_{dd.Br_2\left(tăng\right)}=m_{C_2H_2}=2,6g\)

\(n_{hh}=\dfrac{5,6}{22,4}=0,25mol\)

\(n_{C_2H_2}=\dfrac{2,6}{26}=0,1mol\)

\(\%V_{C_2H_2}=\dfrac{0,1}{0,25}.100=40\%\)

\(\%V_{CH_4}=100\%-40\%=60\%\)

b.\(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)

    0,1         0,2                          ( mol )

\(C_{M\left(dd.Br_2\right)}=\dfrac{0,2}{0,1}=2M\)

Bài 1:

\(n_{HCl}=2.0,16=0,32\left(mol\right);n_{H_2}=\dfrac{3,584}{22,4}=0,16\left(mol\right)\)

PTHH: Mg + 2HCl → MgCl₂ + H₂

PTHH: Fe + 2HCl → FeCl₂ + H₂

\(m_{H_2}=0,16.2=0,32\left(g\right)\)

\(m_{HCl}=0,32.36,5=11,68\left(g\right)\)

Theo ĐLBTKL ta có: \(m_{MgCl_2+FeCl_2}=1,4+11,68-0,32=12,76\left(g\right)\)

Bài 12:

Theo ĐLBTKL, ta có:

\(m_{hhkl}+m_{O_2}=m_{hh.oxit}\\ \Leftrightarrow11,9+m_{O_2}=18,3\\ \Leftrightarrow m_{O_2}=18,3-11,9=6,4\left(g\right)\\ n_{O_2}=\dfrac{6,4}{32}=0,2\left(mol\right)\\ V_{O_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\)