Đặt :

nFe = x mol

nAl = y mol

Fe + 2HCl --> FeCl2 + H2

x____________x______x

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

y______________y______1.5y

mhh= 56x + 27y = 5.5 g (1)

nH2 = x + 1.5y = 0.2 (2)

Giải (1) và (2) :

x = 0.05

y= 0.1

mFe = 2.8 g

mAl = 2.7 g

%Fe = 50.91%

%Al = 49.09%

Từ PTHH ta thấy :

nHCl = 2nH2 = 0.2*2=0.4 mol

mHCl = 0.4*36.5 = 14.6 g

mddHCl = 14.6*100/14.6 = 100g

mdd sau phản ứng = 5.5 + 100 - 0.4 = 105.1 g

mFeCl2 = 6.35g

mAlCl3 = 13.35g

C%FeCl2 = 6.04%

%AlCl3 = 12.7%

Fe + 2HCl → FeCl₂ + H₂ (1)

2Al + 6HCl → 2AlCl₃ + 3H₂ (2)

\(n_{H_2}=\frac{4,48}{22,4}=0,2\left(mol\right)\)

a) Gọi x,y lần lượt là số mol của Fe và Al

Ta có: \(56x+27y=5,5\) (*)

Theo Pt1: \(n_{H_2}=n_{Fe}=x\left(mol\right)\)

Theo PT2: \(n_{H_2}=\frac{3}{2}n_{Al}=1,5y\left(mol\right)\)

Ta có: \(x+1,5y=0,2\) (**)

Từ (*)(**) ta có: \(\left\{{}\begin{matrix}56x+27y=5,5\\x+1,5y=0,2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,05\\y=0,1\end{matrix}\right.\)

Vậy \(n_{Fe}=0,05\left(mol\right)\Rightarrow m_{Fe}=0,05\times56=2,8\left(g\right)\)

\(n_{Al}=0,1\left(mol\right)\Rightarrow m_{Al}=0,1\times27=2,7\left(g\right)\)

\(\%m_{Fe}=\frac{2,8}{5,5}\times100\%=50,91\%\)

\(\%m_{Al}=100\%-50,91\%=49,09\%\)

b) Theo Pt1: \(n_{HCl}=2n_{Fe}=2\times0,05=0,1\left(mol\right)\)

Theo Pt2: \(n_{HCl}=3n_{Al}=3\times0,1=0,3\left(mol\right)\)

\(\Rightarrow\Sigma n_{HCl}=0,1+0,3=0,4\left(mol\right)\)

\(\Rightarrow\Sigma m_{HCl}=0,4\times36,5=14,6\left(g\right)\)

\(\Rightarrow m_{ddHCl}=\frac{14,6}{14,6\%}=100\left(g\right)\)

c) \(m_{H_2}=0,2\times2=0,4\left(g\right)\)

Ta có: \(m_{dd}saupứ=5,5+100-0,4=105,1\left(g\right)\)

Theo Pt1: \(n_{FeCl_2}=n_{Fe}=0,05\left(mol\right)\)

\(\Rightarrow m_{FeCl_2}=0,05\times127=6,35\left(g\right)\)

\(\Rightarrow C\%_{FeCl_2}=\frac{6,35}{105,1}\times100\%=6,04\%\)

Theo Pt2: \(n_{AlCl_3}=n_{Al}=0,1\left(mol\right)\)

\(\Rightarrow m_{AlCl_3}=0,1\times133,5=13,35\left(g\right)\)

\(\Rightarrow C\%_{AlCl_3}=\frac{13,35}{105,1}\times100\%=12,7\%\)

a) \(n_{H_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)

PTHH: 2Al + 6HCl → 2AlCl₃ + 3H₂

Mol:       x                                  1,5x

PTHH: Zn + 2HCl → ZnCl₂ + H₂

Mol:       y                                 y

Ta có: \(\left\{{}\begin{matrix}27x+65y=24,9\\1,5x+y=0,6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}27x+65.\left(0,6-1,5x\right)=24,9\\y=0,6-1,5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,2\\y=0,3\end{matrix}\right.\)

b, \(m_{Al}=0,2.27=5,4\left(g\right);m_{Zn}=24,9-5,4=19,5\left(g\right)\)

c) \(\%m_{Al}=\dfrac{5,4.100\%}{24,9}=21,69\%;\%m_{Zn}=100\%-21,69\%=78,31\%\)

d) 

PTHH: 2Al + 6HCl → 2AlCl₃ + 3H₂

Mol:      0,2     0,6          0,2       0,3

PTHH: Zn + 2HCl → ZnCl₂ + H₂

Mol:      0,3     0,6        0,3       0,3

\(m_{ddHCl}=\dfrac{\left(0,6+0,6\right).36,5.100}{14}=312,857\left(g\right)\)

e) m_{dd sau pứ} = 24,9 + 312,857 - (0,3+0,3).2 = 336,557 (g)

\(C\%_{ddAlCl_3}=\dfrac{0,2.133,5.100\%}{336,557}=7,93\%\)

\(C\%_{ddZnCl_2}=\dfrac{0,3.136.100\%}{336,557}=12,12\%\)

n_H2 = \(\dfrac{4,48}{22,4}\) = 0,2 (mol)

Zn + 2HCl ----> ZnCl₂ + H₂ (1)

ZnO + 2HCl ----> ZnCl₂ + H₂O (2)

n_Zn = n_{ZnCl2 (1)} = n_H2 = 0,2 (mol)

=> m_Zn = 0.2 x 65 = 13 (g)

=> m_ZnO = 21,1 - 13 = 8,1 (g)

=> n_ZnO = 8,1/81 = 0.1 (mol)

n_{ZnCl2 (2)} = n_ZnO = 0.1 (mol)

C%_ZnCl2 = \(\dfrac{152\left(0,2+0,1\right)}{21,1+200}\times100\%=20.62\%\)

Câu 1:

Đặt \(n_{Fe}=x\left(mol\right);n_{Al}=y\left(mol\right)\)

\(n_{HCl}=0,4\left(mol\right)\)

\(Fe^o\rightarrow Fe^{+2}+2e\)

x_____________2x_(mol)

\(Al^o\rightarrow Al^{+3}+3e\)

y____________3y_(mol)

\(2H^-\rightarrow H_2^o+2e\)

0,8_____0,4____0,8_(mol)
\(BTe:2x+3y=0,8\)

Theo đề ta có hệ: \(\left\{{}\begin{matrix}56x+27y=11\\2x+3y=0,8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\%m_{Fe}=\frac{0,1.56}{11}.100\%=51\left(\%\right)\\\%m_{Al}=100-51=49\left(\%\right)\end{matrix}\right.\)

\(BTNT:\Rightarrow\left\{{}\begin{matrix}n_{FeCl_2}=0,1\left(mol\right)\\n_{AlCl_3}=0,2\left(mol\right)\end{matrix}\right.\)

\(m_{hh}=0,1.127+133,5.0,2=39,4\left(g\right)\)

\(m_{ddHCl}=\frac{36,5.0,8.100}{7,3}=400\left(g\right)\)

Câu 2:

\(n_{H_2}=0,15\left(mol\right)\)

\(PTHH:2Al+6HCl\rightarrow2AlCl_3+3H_2\)

(mol)____0,1____0,3______0,1______0,15__

\(\%m_{Al_2O_3}=\frac{7,8-27.0,1}{7,8}.100\%=65,4\left(\%\right)\)

Câu 3:

\(n_{H_2}=0,1\left(mol\right)\)

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

(mol)_____0,1__________________0,1__

\(\%m_{ZnO}=\frac{10,55-0,1.65}{10,55}.100\%=38,4\left(\%\right)\)

Câu 29.

a)Dùng quỳ tím ẩm:

   +Hóa xanh: \(NaOH;Ca\left(OH\right)_2\)

     Nhỏ 1 lượng \(Na_2CO_3\) vào hai chất trên:

     Xuất hiện kết tủa: \(Ca\left(OH\right)_2\)

       \(Ca\left(OH\right)_2+Na_2CO_3\rightarrow CaCO_3\downarrow+2NaOH\)

     Không hiện tượng:\(NaOH\)

    +Qùy không đổi màu: \(NaCl;NaNO_3\)

      Nhỏ 1 ít bạc nitrat \(AgNO_3\) xuất hiện kết tủa: \(NaCl\)

       \(AgNO_3+NaCl\rightarrow AgCl\downarrow+NaNO_3\)

      Không hiện tượng:\(NaNO_3\)

Câu 30.

\(n_{\uparrow}=\dfrac{1,12}{22,4}=0,05mol\)

a)  \(ZnO+2HCl\rightarrow ZnCl_2+H_2O\)

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

     0,05     0,1                     0,05

\(m_{Zn}=0,05\cdot65=3,25\left(g\right)\)\(\Rightarrow\%m_{Zn}=\dfrac{3,25}{30}\cdot100\%=10,83\%\)

\(\Rightarrow m_{ZnO}=100\%-10,83\%=89,17\%\)

b)\(m_{ZnO}=30-3,25=26,75\left(g\right)\)

   \(\Rightarrow n_{ZnO}=\dfrac{26,75}{81}=0,33mol\)

  \(\Sigma n_{HCl}=0,33\cdot2+0,05\cdot2=0,76mol\)

  \(\Rightarrow m_{HCl}=0,76\cdot36,5=27,74\left(g\right)\)

 \(\Rightarrow m_{ddsau}=\dfrac{27,74}{7,4}\cdot100=374,86\left(g\right)\)

1)nH2= 0,224/22,4=0,01(mol)

Mg +2HCl -> MgCl2 + H2

0,01<----------------- 0,01

mMg= 0,01x 24= 0,24(g)

mCu= 0,56- 0,24= 0,32(g)

%Mg = 0,24/0.56x 100% = 42,86%

%Cu = 0,32/0,56 x 100% = 57,14%

2) nHCl = 0,6 x 1 = 0,6 (mol)

Zn + 2HCl -> ZnCl2 + H2

0,2<- 0,4<- 0,2 <- 0,2

nHCl = 0,6- 0,4 = 0,2 (mol)

ZnO + 2HCl -> ZnCl2 + H2O

0,1 <- 0,2

=> mZn = 0,2 x 65 = 13(g)

mZnO = 0,1 x 81 = 8,1 (g)

3) gọi x,y lần lượt là số mol của KCl và NaCl (x,y >0)

KCl + AgNO3 -> AgCl + KNO3

x-----> x---------> x----------> x

NaCl + AgNO3 -> AgCl + NaNO3

y -------> y ----------> y -----> y

ta có hpt

74,5x + 58,5y = 26,6

143,5x + 143,5y = 57,4

=> x=0,2 , y = 0,2

mKCl= 74,5 x 0,2 =14,9 (g)

mNaCl = 58,5 x 0,2 =11,7(g)

%KCl = 14,9/26,6 x 100% =56,02 %

%NaCl= 11,7/26,6 x 100% =43,98%

Mg +2HCl\(\rightarrow\)MgCl2+ H2

MgO+2HCl\(\rightarrow\)MgCl2+H2O

ta có nH2 =\(\frac{2,24}{22,4}\)=0,1 mol

Theo pthh 1 :nH2=nMg =0,1 mol

\(\rightarrow\)mMg=0,1.24 =2,4 g

\(\rightarrow\)mMgO =6-2,4=3,6g

nMgO =\(\frac{3,6}{40}\)=0,09 mol

%mMgO =\(\frac{3,6}{6}.100\%\)=60%

Ta co nHCl =2nMg =0,2 mol

nHCl =2nMgO =0,18 mol

mHCl=(0,18+0,2).36,5=13,87 g

VHCl=\(\frac{13,87}{1}.1\)=12,6ml

Zn + 2HCl → ZnCl₂ + H₂ (1)

ZnO + 2HCl → ZnCl₂ + H₂ (2)

\(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

Gọi \(x\) là số mol Zn

\(\Rightarrow n_{ZnO}=2x\left(mol\right)\)

Theo Pt1: \(n_{H_2}=n_{Zn}=x\left(mol\right)\)

Theo PT2: \(n_{H_2}=n_{ZnO}=2x\left(mol\right)\)

\(\Rightarrow\Sigma n_{H_2}=x+2x=0,15\)

\(\Leftrightarrow3x=0,15\)

\(\Leftrightarrow x=0,05\left(mol\right)\)

Vậy \(n_{Zn}=0,05\left(mol\right)\Rightarrow m_{Zn}=0,05\times65=3,25\left(g\right)\)

\(n_{ZnO}=2\times0,05=0,1\left(mol\right)\Rightarrow m_{ZnO}=0,1\times81=8,1\left(g\right)\)