Por favor ayudenme con estos nueve ejercicios .
Respuestas a la pregunta
Se tienen un conjunto de ecuaciones químicas y se pide realizar balanceo por el método de coeficientes indeterminados.
1. NaCl + H₂SO₄ + O₂ → Na₂SO₄ + H₂O + Cl₂
Colocamos letras como coeficientes:
a NaCl + b H₂SO₄ + c O₂ → d Na₂SO₄ + e H₂O + f Cl₂
Na a = 2d asumiendo a = 1 …… d = 1/2
Cl a = 2f 2f = 1 ……. f = 1/2
H 2b = 2e 2e = 2 (1/2) ……. e = 1/2
S b = d b = 1/2
O 4b + 2c = 4d + e 4(1/2) + 2c = 4(1/2) + 1/2 … 2 + 2c = 5/2 ... c = 1/4
Colocamos en la ecuación los coeficientes calculados:
NaCl + 1/2 H₂SO4 + 1/4 O₂ → 1/2 Na₂SO₄ + 1/2 H₂O + 1/2 Cl₂
Multiplicando x 4 :
4 NaCl + 2 H₂SO4 + O₂ → 2 Na₂SO₄ + 2 H₂O + 2 Cl₂
2. H₂SO₄ + HCl → SO₂ + H₂O + Cl₂
a H₂SO₄ + b HCl → c SO₂ + d H₂O + e Cl₂
H: 2a + b = 2d 2 (1) + b = 2(2) ……… b = 2
S: a = c asumiendo a = 1 ….. c = 1
O: 4a = 2c + d 4(1) = 2(1) + d ………. d = 2
Cl: b = 2e 2 = 2e …….. e = 1
H₂SO4 + 2 HCl → SO₂ + 2 H₂O + Cl₂
3. NaOH + Cl₂ → NaCl + NaClO₃ + H₂O
a NaOH + b Cl₂ → c NaCl + d NaClO₃ + e H₂O
Na: a = c + d 1 = c + 1/6 ------ c = 5/6
O: a = 3d + e 1 = 3d + 1/2 ----- d = 1/6
Cl: 2b = c + d b = (5/6 + 1/6)/2 b = 1/2
H: a = 2 e Asumimos a = 1 , entonces e = 1/2
NaOH + 1/2 Cl₂ → 5/6 NaCl + 1/6 NaClO₃ + 1/2 H₂O
Multiplicando por 6:
6 NaOH + 3 Cl₂ → 5 NaCl + NaClO₃ + 3 H₂O
4. As₂O₃ + H₂SO₄ + Zn → AsH₃ + H₂O + ZnSO₄
a As₂O₃ + b H₂SO₄ + c Zn → d AsH₃ + e H₂O + f ZnSO₄
As: 2a = d Asumiendo a = 1 …….. d = 2
O: 3a + 4b = e + 4f 3(1) + 4b = e + 4b …….. e = 3
H: 2b = 3d + 2e 2b = 3(2) + 2(3) ….. 2b = 6 + 6 ……. b = 6
S: b = f f= 6
Zn: c = f c = 6
As₂O₃ + 6 H₂SO₄ + 6 Zn → 2 AsH₃ + 3 H₂O + 6 ZnSO₄
5. PbO₂ + HCl → PbCl₂ + Cl₂ + H₂O
a PbO₂ + b HCl → c PbCl₂ + d Cl₂ + e H₂O
Pb: a = c Asumiendo a = 1 ….. c = 1
O : 2a = e e = 2
H: b = 2e b = 4
Cl : b = 2c + 2d 4 = 2 + 2d ------ d = 1
PbO₂ + 4 HCl → PbCl₂ + Cl₂ + 2 H₂O
6. I₂ + HNO₃ → HIO₃ + NO + H₂O
a I₂ + b HNO₃ → c HIO₃ + d NO + e H₂O
I : 2a = c asumiendo a = 1 ……. c = 2
H : b = c + 2e b = 2 + 2e ……. b = 10/3
N : b = d d = 2 + 2e … d = 2 + 2 (2/3) … d = 10/3
O : 3b = 3c + d + e 3(2 + 2e) = 3(2) + (2 + 2e) + e …… e = 2/3
I₂ + 10/3 HNO₃ → 2 HIO₃ + 10/3 NO + 2/3 H₂O
Multiplicando x 3 :
3 I₂ + 10 HNO₃ → 6 HIO₃ + 10 NO + 2 H₂O
7. K₂Cr₂O₇ + SnCl₂ + HCl → CrCl₃ + SnCl₄ + KCl + H₂O
a K₂Cr₂O₇ + b SnCl₂ + c HCl → d CrCl₃ + e SnCl₄ + f KCl + g H₂O
K: 2a = f Asumiendo a = 1 ....... f = 2
Cr: 2a = d d = 2(1) d = 2
O: 7a = g g = 7(1) g = 7
Sn: b = e b = 3
Cl: 2b + c = 3d + 4e + f ....... 6 + 14 = 6 + 4e + 2 ..... e = 3
H: c = 2g c = 2 (7) c = 14
K₂Cr₂O₇ + 3 SnCl₂ + 14 HCl → 2 CrCl₃ + 3 SnCl₄ + 2 KCl + 7 H₂O
8. CuS + HNO₃ → Cu (NO₃)₂ + S + H₂O + NO
a CuS + b HNO₃ → c Cu (NO₃)₂ + d S + e H₂O + f NO
Cu: a = c Asumiendo: a = 1 ------ c = 1
S: a = d d = 1
H: b = 2e b = 8/3
N: b = 2c + f 2e = 2 + f ----- f = 2e – 2 ------ f = 2/3
O: 3b = 6c + e + f 6e = 6 + e + f --- 5e = 6 + 2e – 2 -- e = 4/3
CuS + 8/3 HNO₃ → Cu (NO₃)₂ + S + 4/3 H₂O + 2/3 NO
Multiplicando x 3 :
3 CuS + 8 HNO₃ → 3 Cu (NO₃)₂ + 3 S + 4 H₂O + 2 NO
9. Zn₃Sb₂ + H₂O → Zn (OH)₂ + H₃Sb
a Zn₃Sb₂ + b H₂O → c Zn (OH)₂ + d H₃Sb
Zn : 3a = c asumiendo a = 1 … c = 3
Sb : 2a = d 2 (1) = d d = 2
H : 2b = 2c + 3d
O : b = 2c b = 2 (3) b = 6
Zn₃Sb₂ + 6 H₂O → 3 Zn (OH)₂ + 2 H₃Sb