Clasificacion

 Clasificacion

1/10I2O5 + BrF3 -> 1/5IF5 + 1/4O2 + BrF 2 =  Descomposición

1/2P4O10 + 3Mg(OH)2 -> Mg3(PO)2 + 3H2O = Neutralización

1(NH4)2 Cr2O7 -> Cr2O3 + N2 + 4H2O = Doble desplazamiento                                                                                                                                                                                                                                                                                                                                                                                                                              

Balance

Balance

(NH4)2 Cr2O7-> Cr2O3 + N2 + H2O

N = 2                      Cr = 2

H = 8                      O = 3 + 1

Cr = 2                     N =  2

O = 7                      H = 2

Para N             Para H

   

 2a = 2c            8 a = 2d

Para Cr           Para O

2 a = 2b         7 a = 3b + d

a =1

b =1

c = 1

d =4

1(NH4)2 Cr2O7 -> Cr2O3 + N2 + 4H2O

N = 2                      Cr = 2

H = 8                      O = 3 + 4

Cr = 2                    N =  2

O = 7                      H = 8

P4O10 + Mg(OH)2 -> MG3(PO4)2 + H2O

P = 4                                  P = 2

O = 10 + 2                        O = 8 + 1

Mg = 1                              Mg = 3

H = 2                                  H = 2 

Para P            Para O

 4a = 2c            10 a + 2b = 8c + d

Para H           Para Mg

2 b = 2d         b = 3c

a = 1/2

b = 3

c = 1

d = 3

1/2P4O10 + 3Mg(OH)2 -> Mg3(PO)2 + 3H2O

P = ½ * 4 = 2                                 P = 2

O = ½ * 10 = 5 + 6 = 11             O = 8 + 3 = 11

Mg = 3                                           Mg = 3

H = 2 * 3 = 6                                 H = 2 * 3 = 6 

 I2O5 + BrF3 -> IF5 + O2 + BrF 2

I = 2                                 I = 2

O = 5                               F = 7

Br = 1                               O  = 2

F = 3                                 Br = 1

Para I            Para O

 2a = c            5 a = 2d

Para Br           Para F

b = e                3b = 5c + 2e

a = 1/10

b = 1

c = 1/5

d = 1/4

e = 1      

1/10I2O5 + BrF3 -> 1/5IF5 + 1/4O2 + BrF 2  

I = 2 * 1/10 = 1/5              I = 1/5

O = 5 * 1/10 = 1/2             F = 1/5 * 5 = 1 + 2 = 3

Br = 1                               O  = ¼ * 2 = ½

F = 3                                 Br

Analisis cuatitativo

Análisis cuantitativo

# de átomos	N=2    O=7 H=8 Cr=2	Cr=2 O=3	N=2	H=2 O=1
# de moléculas	1(NH4)2Cr2O7	1Cr2O3	1N2	4H2O
# de moles	1(NH4)2Cr2O7	1Cr2O3	1N2	4H2O
# de gramos	252 gr	152 gr	28 gr	72 gr

# de átomos	P=4 O=10	Mg=3 O=2 H=2	Mg=3 P=2 O=8	H=2 O=1
# de moléculas	1/2P4O10	3Mg(OH)2	1Mg3(PO4)2	3H2O
# de moles	1/2P4O10	3Mg(OH)2	1Mg3(PO4)2	3H2O
# de gramos	140.5 gr	174 gr	262 gr	22 gr

# de átomos	I=2 O=5	Br=1 F=3	I=1 F=5	O=2	Br=1 F=2
# de moléculas	1/10 I2O5	1BrF3	1/5IF5	1/4O2	1BrF2
# de moles	1/10 I2O5	1BrF3	1/5IF5	1/4O2	1BrF2
# de gramos	33.4 gr	137 gr	44.4 gr	8 gr	118 gr

Tarea pág. 125 no. 4.53, 4.55, 4.56, 4.59

P.125 nr. 4.53
4.53.:
KI= Yoduro de Potasio
# de gramos = 2.80M x (5.00x100mL)x 166
# de gramos = 2.80M x 0.5 lt x 166
# de gramos = 232.4 gr

4.55
MgCl2-> Cloruro de Magnesio (II)

Molaridad = # de moles / 1 lt de sal

0.100M = # moles / 0.6 lt
# de moles = 0.6 lt / 0.100M
# de moles = 6 moles

4.56

# gramos = 5.50M x 0.035 mL x 56
# gramos = 107.8
= 10.78gr

4.59
A. NaCl
# de gramos = M x V x Pm
2.14g = 0.270 x 58
2.14g / (0.270 x 58) = V
136.6 lt = V

B.Etanol

4.30gr / (1.50 x 46) = V
0.0623 ml = V
623 lt = V

C. Ácido acético

0.85gr / (0.30 x 60) = V
0.0472mL = V
472 lt = V