Ta có: \(\left\{{}\begin{matrix}n_{H_2SO_4}=\dfrac{0,098}{98}=0,001\left(mol\right)\\n_{HNO_3}=0,2.0,04=0,008\left(mol\right)\end{matrix}\right.\)

=> \(\sum n_{H^+}=2n_{H_2SO_4}+n_{HNO_3}=0,01\left(mol\right)\)

=> \(\left[H^+\right]=\dfrac{0,01}{0,5+0,2}=\dfrac{1}{70}M\)

=> \(pH=-\log\left(\dfrac{1}{70}\right)=1,845\)

Câu 4:

1) \(m_{Pb}=n.M=0,64.207=132,48\left(g\right)\)

2) \(m_{KNO_3}=n.M=1,3.101=131,3\left(g\right)\)

3) \(m_{CO_2}=n.M=0,075.44=3,3\left(g\right)\)

4) \(n_{CH_4}=\dfrac{V_{\left(đktc\right)}}{22,4}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

=> \(m_{CH_4}=n.M=0,1.16=1,6\left(g\right)\)

do hai nguyên tố hóa học cacbon và hidro tạo nên

có 2 nguyên tử C, 4 nguyên tử H trong 1 phân tử

PTK(C2H4):12*2+1*4=28(dvC)

1,n=\(\dfrac{m}{M}\)=\(\dfrac{0,96}{64}\)=0,015(mol)

3,n=\(\dfrac{m}{M}=\dfrac{70}{56}=1,25\left(mol\right)\)

9,n\(=\dfrac{V}{22,4}=\dfrac{8,064}{22,4}=0,36\left(mol\right)\)

1) \(n_{Cu}=\dfrac{m}{M}=\dfrac{0,96}{64}=0,015\left(mol\right)\)

2) \(n_{Ba}=\dfrac{m}{M}=\dfrac{30,14}{137}=0,22\left(mol\right)\)

3) \(n_{CaO}=\dfrac{m}{M}=\dfrac{70}{40+16}=1,25\left(mol\right)\)

4) \(n_{NH_3}=\dfrac{m}{M}=\dfrac{6,12}{14+3}=0,36\left(mol\right)\)

5) \(n_{H_2SO_4}=\dfrac{m}{M}=\dfrac{37,24}{2+32+16\cdot4}=0,38\left(mol\right)\)

6) \(n_{HNO_3}=\dfrac{m}{M}=\dfrac{157,5}{1+14+16\cdot3}=2,5\left(mol\right)\)

7) \(n_{Al_2\left(SO_4\right)_3}=\dfrac{m}{M}=\dfrac{25,65}{27\cdot2+\left(32+16\cdot4\right)\cdot3}=0,075\left(mol\right)\)

8) \(n_{BaCO_3}=\dfrac{m}{M}=\dfrac{76,83}{137+12+16\cdot3}=0,39\left(mol\right)\)

9) \(n_{NH_3\left(dktc\right)}=\dfrac{V}{22,4}=\dfrac{8,064}{22,4}=0,36\left(mol\right)\)

10) \(n_{N_2O\left(dktc\right)}=\dfrac{V}{22,4}=\dfrac{41,664}{22,4}=1,86\left(mol\right)\)

a) \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

PTHH: Fe + 2HCl ---> FeCl₂ + H₂

           0,1<--------------0,1<----0,1

=> \(\left\{{}\begin{matrix}m_{Fe}=0,1.56=5,6\left(g\right)\\m_{Cu}=8-5,6=2,4\left(g\right)\end{matrix}\right.\)

b) \(C_{M\left(FeCl_2\right)}=\dfrac{0,1}{0,2}=0,5M\)

a) \(4Na+O_2->2Na_2O\)

tỉ lê: 4:1:2

b) \(4Al+3O_2->2Al_2O_3\)

tỉ lệ: 4:3:2

c) \(2Fe+3Cl_2->2FeCl_3\)

tỉ lệ: 2:3:2

d) \(P_2O_5+2H_2O->2H_3PO_4\)

tỉ lệ: 1:2:2

Câu 1:

(a) \(4Na+O_2\xrightarrow[]{t^o}2Na_2O\)

Tỉ lệ số nguyên tử Na : số phân tử O₂ : số phân tử Na₂O = 4 : 1 : 2

(b) \(4Al+3O_2\xrightarrow[]{t^o}2Al_2O_3\)

Tỉ lệ số nguyên tử Al : số phân tử O₂ : số phân tử Al₂O₃ = 4 : 3 : 2

(c) \(2Fe+3Cl_2\xrightarrow[]{t^o}2FeCl_3\)

Tỉ lệ số nguyên tử Fe : số phân tử Cl₂ : số phân tử FeCl₃ = 2 : 3 : 2

(d) \(P_2O_5+3H_2O\rightarrow2H_3PO_4\)

Tỉ lệ số phân tử P₂O₅ : số phân tử H₂O : số phân tử H₃PO₄ = 1 : 3 : 2

n=\(\dfrac{m}{M}\)=\(\dfrac{0,64}{32}\)=0,02(mol)

\(n_{O_2}=\dfrac{m}{M}=\dfrac{0,6}{32}=0,02\left(mol\right)\)

1) \(Na_2O+H_2O->2NaOH\)

2) \(P_2O_5+3H_2O->2H_3PO_4\)

3) \(O_2+2H_2-^{t^o}>2H_2O\)

4) \(2KClO_3->2KCl+3O_2\)

5) \(2Mg+O_2->2MgO\)

6) \(C+O_2->CO_2\)

a, 4Na+O\(_2\)→2Na\(_2\)O

b,4Al+3O\(_2\)→2Al\(_2O_3\)

c,2Fe+3Cl\(_2\)→2FeCl\(_3\)

Câu 1:

$\rm (a) 4Na + O_2 \xrightarrow{t^o} 2Na_2O (4:1:2)$
$\rm (b) 4Al + 3O_2 \xrightarrow{t^o} 2Al_2O_3 (4:3:2)$
$\rm (c) 2Fe + 3Cl_2 \xrightarrow{t^o} 2FeCl_3 (2:3:2)$
$\rm (d) P_2O_5 + 3H_2O \rightarrow 2H_3PO_4 (1:3:2)$
$\rm (e) 2Fe(OH)_3 \xrightarrow{t^o} Fe_2O_3 + 3H_2O (2:1:3)$
$\rm (f) 2KClO_3 \xrightarrow[MnO_2]{t^o} 2KCl + 3O_2 (2:2:3)$
$\rm (g) Mg + 2HCl \rightarrow MgCl_2 + H_2 (1:2:1:1)$
$\rm (h) Fe_2O_3 + 6HCl \rightarrow 2FeCl_3 + 3H_2O (1:6:2:3)$
$\rm (i) 2Al + 6HCl \rightarrow 2AlCl_3 + 3H_2 (2:6:2:3)$
$\rm (j) 2C_2H_6 + 7O_2 \xrightarrow{t^o} 4CO_2 + 6H_2O (2:7:4:6)$
$\rm (k) BaCl_2 + 2AgNO_3 \rightarrow Ba(NO_3)_2 + 2AgCl (1:2:1:2)$

$\rm (m) Al_2(SO_4)_3 + 3Ba(OH)_2 \rightarrow 2Al(OH)_3 + 3BaSO_4 (2:3:2:3)$
$\rm (n) Cu + 2H_2SO_{4(đ)} \xrightarrow{t^o} CuSO_4 + SO_2 + 2H_2O (1:2:1:1:2)$