Tarea quimica 5

                                               Tarea química 5.

Ejercicios de la hoja 3, 5 y 6.

Análisis cuantitativo:

3)                                   (NH4)2 Cr2 O7               →              Cr2 O3+ N2+ H2O

#  Átomos             N= 1  H=8     Cr= 2  O=7                Cr= 2 O= 3 N=2 H= 2 O=1

# Moléculas          1 NHCrO 7                 1CrO3                1N2                      1H2O               

# Moles                  1 NHCrO 7                1CrO3                 1N2                     1H2O

Gramos                  179.01 gr                  100 gr                 28 gr            33.008 gr

                                                                                                   = 161.008 gr 

Clasificación y balance:

a (NH4)2 Cr2 O7       →    b Cr2 O3 +  c N2  +d  H2O=  Descomposición.

N= 2                                  Cr= 2       N= 2     H= 2

H= 8                                 O= 3                       O= 1

Cr= 2

O= 7

Para N

2a= c                                                           a= 1

Para H                                                         b= 2

3a + b= 2c + d                                            c= 2

Para  Cr                                                       d=  2/3

2a= 2d

Para O   b= 2c

5)                                 P4 O10 +  Mg(OH)2           →  Mg3(PO4)2 +H2O

# Átomos           P=4    O=10    Mg=1  O=2 H=2       Mg=3 P=2 O=8  H=2 O=1

# Moléculas       1PO10            1Mg(OH)2              1Mg(PO4)2         1H2O  

# Moles               1PO10            1Mg(OH)2             1Mg(PO4)2          1H2O

Gramos              190.97gr         42.326gr                 183.28 gr             18.016gr

a P4 O10+ b Mg (OH)2  →c Mg3 (PO4)2 + d H2O= Neutralización.

P=4                Mg=1                Mg=3                   H=2

O=10            O=2                    P=2                      O=1

                      H=2                  O=8

Para P

4a=c             1=4a

                    ¼=a

Para O

10a+ b= 2c + d                                          a= 1/4

Para Mg                                                     b= 6

B= 3c + 2d                                                 c= 3

B= 6                                                            d= 1/5

Para H

B= 3c

6)                          I2 O5    + BrF3                → I F5 +       O2    + BrF2               

# átomos        I= 2  O=5     Br=1  F=3               I=1  F=5   O=2       Br=1   F=2

#moléculas      1IO5             1BrF3                    1IF5            1O2         1BrF2

# Moles          1IO5               1BrF3                     1IF5            1O2        1BrF2

Gramos          206.9gr         136.9gr                  221.9gr       32gr         117.9gr

I2 O5+ Br F3 → IF5+ O2+ BrF2 =  Desplazamiento

Tarea quimica 4

                                                  Tarea química 4.

Pagina 86:

3.39-    Sn- estaño                O- 16

              SnO2                       Sn- 118

                                               Pm- 134

% elemento=   masa del elemento/ Pm  *100

% Sn=   118q/134q   * 100

% Sn= 88.06

% O= 16*2/134  * 100

%= 23.881

Sn= 88.06%

O= 23.882%

3.41-      Ca H10 O                                                              C= 12*9= 108

                                                                                              H= 10

                                                                                               O= 16

a) C= 80.597%           H= 7.463%                                        O= 11.940%

    % C= 108/134  *100

    %C= 80.597

%H = 10/134  *100                                                       %O= 16/134  *100

%H= 7.463                                                                       %O= 11.940

  3.43-       a)        C= 44.4%                                        O= 9.86%

                           H= 6.21%

                           S= 39.5%                   

1) Convertir a moles:

44.4 g 1mol c/ 12g=   3.7 mol C

6.21g    1 mol H/1g= 6.21 mol H

39.5g     1 mol S/ 32g= 1.234 mol S

9.86g    1 mol O/ 16g= 0.616 mol O

Encontrar factor común:

C= 3.7 mol C/0.616 mol= 6.006 C

H= 6.21 mol H/ 0.616 mol= 10.08 H                        =            C6 H10 S2 0

S= 1.234 mol S/ 0.616 mol= 2.003 S

O= 0.616 mol O/ 0.616 mol= 1

b)  162g/ 162g= 1               =                     C6 H S2 O

Libro pagina 125:

4.53-  # gramos= M     V     Pm

            #gramos= (2.80)     (0.500 lt)     (87)

           #gramos= 121.8 gramos es la masa de IK.

4.55-  MaCl2 =  Cloruro de magnesio

                                      60.0 ml

                                      0.100 M

Ma= 24

Cl= 35*2=70

                 94 Pm                 

            Molaridad= # moles/ 1 lt sal

0.100 M= # moles/ 0.6 lt

0.100/ # moles= 0.6 H

Moles= 0.6/ 0.100 M=  # moles= 6

4.56:   Ko lt

            35.0 ml

            5.50 M

K= 39

O= 16              =56 Pm                            # gramos= 5.50 M     0.035 ml  * 56

H= 1                                                          #gramos= 107.8      

                                                                        107.8= 10.78 gr

4.59-  V= ?                                      Na= 23

                                                        Cl= 35                       =58

a) 2.14- NaCl

    0.270 M

                           # gramos= M  *V  *PM

                           #gramos= 0.270 *V * 58

                           2.14 g= 0.270 * V * 58  

                          2.14/58= 0.270 * V

                       2.14/0.270= 136.6 H

b)  4.30 g de estanol                       V=?

          1.50 M                            Estanol= C2 H S O H

C= 12*2= 24

H= 1*6= 6

O= 16= 16

              46 Pm                            430 g/ 1.50* 46= V

 0.0623 ml= V

623 lt= V

c) 0.85 g – acido acético= CH3 COO H

C= 12*2= 24

H= 1*4= 4                     = 60

O= 16*2= 32

0.85g/ 0.30*60= V

0.0472 ml= V

472 lt= V