Bài 39: Luyện tập về tốc độ phản ứng và cân bằng hóa học

\(n_{SO_2} = \dfrac{7,28}{22,4} = 0,325(mol)\\ n_{Al} = a(mol) ; n_{Zn} = b(mol) \Rightarrow 27a + 65b = 10,55(1)\\ \text{Bảo toàn electron :}\\ 3a + 2b = 0,325.2(2)\\ (1)(2) \Rightarrow a = 0,15 ; b = 0,1\\ \%m_{Al} = \dfrac{0,15.27}{10,55}.100\% = 38,39\%\)

1/

\(4FeS_2 + 11O_2 \xrightarrow{t^o} 2Fe_2O_3 + 8SO_2\\ SO_2 + 2H_2S \to 3S + 2H_2O\\ S + H_2 \xrightarrow{t^o,xt} H_2S\\ H_2S + 4Br_2 + 4H_2O \to 8HBr + H_2SO_4\\ H_2SO_4 + NaCl \xrightarrow{t^o} NaHSO_4 + HCl\\ MnO_2 + 4HCl \to MnCl_2 + Cl_2 + 2H_2O\\ 2Na +C l_2 \xrightarrow{t^o} 2NaCl\\ NaCl + AgNO_3 \to AgCl + NaNO_3\)

2/

\(S + H_2 \xrightarrow{t^o} H_2S\\ H_2S + \dfrac{3}{2}O_2 \xrightarrow{t^o} SO_2 + H_2O\\ 2SO_2 +O_2 \xrightarrow{t^o,xt} 2SO_3\\ SO_3 + H_2O \to H_2SO_4\\ CuO + H_2SO_4 \to CuSO_4 + H_2O\\ CuSO_4 + BaCl_2 \to BaSO_4 + CuCl)2\)

1) 

4FeS2 + 11O2 -to-> 2Fe2O3 + 8SO2 

SO2 + 2H2S -to-> 3S + 2H2O 

S + H2 -to-> H2S 

H2S + 4Br2 + 4H2O => H2SO4 + 8HBr 

BaCl2 + H2SO4 => BaSO4 + 2HCl 

2HCl-dp-> H2 + Cl2 

Na + 1/2Cl2 -to-> NaCl 

NaCl + AgNO3 => AgCl + NaNO3 

2) 

H2 + S -to-> H2S 

2H2S + 3O2 -to-> 2SO2 + 2H2O 

SO2 + 1/2O2 -to,V2O5-> SO3

SO3 + H2O => H2SO4 

H2SO4 + CuO => CuSO4 + H2O 

CuSO4 + BaCl2 => BaSO4 + CuCl2 

3) 

MnO2 + 4HCl(đ) -to-> MnCl2 + Cl2 + 2H2O 

Cl2 + H2 -as-> 2HCl 

FeS + 2HCl => FeCl2 + H2S 

2H2S + 3O2 -to-> 2SO2 + 2H2O 

2NaOH + SO2 => Na2SO3 + H2O 

Na2SO3 + H2SO4 => Na2SO4 + SO2 + H2O 

4) 

2KMnO4 + 16HCl(đ) => 2KCl + 2MnCl2 + 5Cl2 +8H2O 

Cl2 + H2 -as-> 2HCl 

Fe(OH)3 + 3HCl => FeCl3 + 3H2O 

FeCl3 + 3AgNO3 => Fe(NO3)3 + 3AgCl 

AgCl -to-> Ag + 1/2Cl2 

2NaBr +  Cl2 => 2NaCl + Br2 

2NaI + Br2 => 2NaBr + I2 

Bài 1 : 

\(n_{Mg}=a\left(mol\right),n_{Zn}=b\left(mol\right)\)

\(m_{hh}=24a+65b=31.5\left(g\right)\left(1\right)\)

\(n_{H_2}=\dfrac{17.92}{22.4}=0.8\left(mol\right)\)

\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)

\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

\(n_{H_2}=a+b=0.8\left(mol\right)\left(2\right)\)

\(\left(1\right),\left(2\right):a=0.5,b=0.3\)

\(m_{Mg}=0.5\cdot24=12\left(g\right)\)

\(m_{Zn}=0.3\cdot19.5\left(g\right)\)

\(n_{H_2SO_4}=n_{H_2}=0.8\left(mol\right)\)

\(C\%H_2SO_4=\dfrac{0.8\cdot98}{300}\cdot100\%=26.13\%\)

Bài 5 :

n Cu = a(mol) ; n Al = b(mol) ; n Mg = c(mol)

=> 64a + 27b + 24c = 11,5(1)

2Al + 6HCl $\to$ 2AlCl3 + 3H2

Mg + 2HCl $\to$ MgCl2 + H2

Theo PTHH :

n H2 = 1,5b + c = 5,6/22,4 = 0,25(2)

Bảo toàn electron :

2n SO2 = 2n Cu 

2.0,1 = 2a (3)

Từ (1)(2)(3) suy ra a = 0,1 ; b = 0,1 ; c = 0,1

Vậy : 

m Cu = 0,1.64 = 6,4 gam

m Al = 0,1.27 = 2,7 gam

m Mg = 0,1.24 = 2,4 gam

Bài 6 :

a) n Mg = a(mol) ; n Fe = b(mol) => 24a + 56b = 6,8(1)

Mg + 2HCl → MgCl2 + H2
Fe + 2HCl → FeCl2 + H2

Theo PTHH : n H2 = a + b = 3,36/22,4 = 0,15(2)

(1)(2) => a = 0,05 ; b = 0,1

%m Mg = 0,05.24/6,8  .100% = 17,65%

%m Fe = 100% -17,65% = 82,35%

b) Bảo toàn e :  

2n Mg + 3n Fe = 2n SO2

=> n SO2 = (0,05.2 + 0,1.3)/2 = 0,2 mol

=> V SO2 = 0,2.22,4 = 4,48(lít)

Bài 7 : 

\(n_{H_2SO_4}=\dfrac{100\cdot10^6\cdot98}{100\cdot98}=10^6\left(mol\right)\)

\(n_{FeS_2}=\dfrac{10^6}{2}=5\cdot10^5\left(mol\right)\)

\(n_{FeS_2\left(tt\right)}=\dfrac{5\cdot10^5\cdot120}{75\%}=80\left(tấn\right)\)

Câu 7 : 

n H2SO4 = 100.1000.98%/98 = 1000(kmol)

Bảo toàn nguyên tố với S :

n FeS2 pư = 1/2 n H2SO4 = 500(kmol)

n FeS2 đã dùng = 500/75% = 2000/3(kmol)

m FeS2 = 120 . 2000/3 = 80 000(kg)

m pirit = 80 000/10% = 800 000(kg)

Bài 8 : 

Coi Oxit sắt gồm Fe(x mol) và O(y mol)

=> 56x + 16y = 36(1)

n SO2 = 5,6/22,4 = 0,25(mol)

Bảo toàn electron : 

3n Fe = 2n SO2 + 2n O

<=> 3a - 2b = 0,5(2)
Từ(1)(2) suy ra a = 0,5 ;  b = 0,5

n Fe / n O = 0,5/0,5 = 1/1

Vậy oxit sắt là FeO

Bài 8 : 

CT : Fe_xO_y

\(n_{SO_2}=\dfrac{5.6}{22.4}=0.25\left(mol\right)\)

\(xFe^{\dfrac{2y}{x}}\rightarrow xFe^{3+}+\left(3x-2y\right)e\)

\(S^{+6}+2e\rightarrow S^{+4}\)

\(BTe:\)

\(n_{Fe_xO_y}=\dfrac{0.5}{3x-2y}\left(mol\right)\)

\(M=\dfrac{36}{\dfrac{0.5}{3x-2y}}=72\left(3x-2y\right)\left(\dfrac{g}{mol}\right)\)

\(\Leftrightarrow56x+16y=72\cdot\left(3x-2y\right)\)

\(\Leftrightarrow160x=160y\)

\(\Leftrightarrow\dfrac{x}{y}=\dfrac{1}{1}\)

\(CT:FeO\)

Câu 9 : 

n S = 3,2/32 = 0,1(mol)

n SO2 = 13,44/22,4 = 0,6(mol)

Bảo toàn e với cả quá trình : 

\(Fe^0 \to Fe^{3+} + 3e\\ S^0 \to S^{+6} + 6e\\ S^{+6} + 2e \to S^{+4}\)

Ta có:  3n Fe + 6n S = 2n SO2

<=> n Fe = (0,6.2 - 0,1.6)/3 = 0,2(mol)

<=> m = 0,2.56 = 11,2(gam)

a) Đặt \(\left\{{}\begin{matrix}n_{Mg}=a\left(mol\right)\\n_{Fe}=b\left(mol\right)\end{matrix}\right.\) \(\Rightarrow24a+56b=22,8\)  (1)

Ta có: \(n_{SO_2}=\dfrac{15,68}{22,4}=0,7\left(mol\right)\)

Bảo toàn electron: \(2a+3b=0,7\cdot2=1,4\)  (2)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}a=0,25\\b=0,3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,25\cdot24}{22,8}\cdot100\%\approx26,32\%\\\%m_{Fe}=73,68\%\end{matrix}\right.\)

b) PTHH: \(2NaOH+H_2SO_{4\left(dư\right)}\rightarrow Na_2SO_4+2H_2O\)

Theo PTHH: \(n_{H_2SO_4\left(dư\right)}=\dfrac{1}{2}n_{NaOH}=\dfrac{0,4\cdot3}{2}=0,6\left(mol\right)\)

Ta có: \(\left\{{}\begin{matrix}n_{Fe_2\left(SO_4\right)_3}=\dfrac{1}{2}n_{Fe}=0,15\left(mol\right)\\n_{MgSO_4}=n_{Mg}=0,25\left(mol\right)\end{matrix}\right.\)

Bảo toàn nguyên tố: \(n_{H_2SO_4\left(p/ứ\right)}=n_{MgSO_4}+3n_{Fe_2\left(SO_4\right)_3}+n_{SO_2}=1,4\left(mol\right)\)

\(\Rightarrow\Sigma n_{H_2SO_4}=1,4+0,6=2\left(mol\right)\) \(\Rightarrow m_{ddH_2SO_4}=\dfrac{2\cdot98}{98\%}=200\left(g\right)\)

Câu 1 : 

\(2KMnO_4+16HCl_{\left(đ\right)}\underrightarrow{^{t^0}}2KCl+2MnCl_2+5Cl_2+8H_2O\)

\(H_2+Cl_2\underrightarrow{^{as}}2HCl\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

\(FeCl_2+2NaOH\rightarrow Fe\left(OH\right)_2+2NaCl\)

Câu 2 : 

\(n_{Mg}=a\left(mol\right),n_{Zn}=b\left(mol\right)\)

\(m_{hh}=24a+65b=14.55\left(g\right)\left(1\right)\)

\(n_{H_2}=\dfrac{7.84}{22.4}=0.35\left(mol\right)\)

\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)

\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

\(n_{H_2}=a+b=0.35\left(mol\right)\left(2\right)\)

\(\left(1\right),\left(2\right):a=0.2,b=0.15\)

\(\%m_{Mg}=\dfrac{0.2\cdot24}{14.55}\cdot100\%=33.98\%\)

\(\%m_{Zn}=100-33.98=67.01\%\)

Câu 3 : 

\(m_{FeS_2}=187.5\cdot80\%=150\left(kg\right)\)

\(n_{FeS_2}=\dfrac{150\cdot10^3}{120}=1250\left(mol\right)\)

\(n_{H_2SO_4}=2n_{FeS_2}=1250\cdot2=2500\left(mol\right)\)

\(n_{H_2SO_4\left(tt\right)}=2500\cdot90\%=2250\left(mol\right)\)

\(m_{dd_{H_2SO_4}}=\dfrac{2250\cdot98\cdot100}{98}=225000\left(g\right)\)

\(V_{dd_{H_2SO_4}}=\dfrac{225000}{1.84}=122282\left(ml\right)=122.282\left(l\right)\)