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12 March, 07:50

Balancing the following equations:

(a) PCl5 (s) + H2O (l) - > POCl3 (l) + HCl (aq)

(b) Cu + HNO3 - > Cu (NO3) 2 + H2O + NO

(c) H2 + I2 - > HI

(d) Fe + O2 - > Fe2O3

(e) Na + H2O - > NaOH + H2

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

(g) P4 + Cl2 - > PCl3

(h) PtCl4 - > Pt + Cl2

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Answers (1)
  1. 12 March, 08:14
    0
    a) PCl5 (s) + H2O (l) ↔ POCl3 (l) + 2HCl (aq)

    b) 3Cu + 8HNO3 ↔ 3Cu (NO3) 2 + 4H2O + 2NO

    c) H2 + I2 ↔ 2HI

    d) 4Fe + 3O2 ↔ 2Fe2O3

    e) 2Na + 2H2O ↔ 2NaOH + H2

    f) (NH4) 2Cr2O7 ↔ Cr2O3 + N2 + 4H2O

    g) P4 + 6Cl2 ↔ 4PCl3

    h) PtCl4 ↔ Pt + 2Cl2

    Explanation:

    a) PCl5 (s) + H2O (l) ↔ POCl3 (l) + 2HCl (aq)

    1 - P - 1

    5 - Cl - 5

    1 - O - 1

    2 - H - 2

    b) 3Cu (0) + 8HN (+5) O3 ↔ 3Cu (+2) (NO3) 2 + 4H2O + 2N (+2) O

    ∴ Cu (0) → Cu (+2) ... in parenthesis the oxidation state

    ⇒ Cu : lose 2e-

    ∴ N (+5) → N (+2)

    ⇒ N : gains 3e-

    3 - Cu - 3

    8 - N - 8

    24 - O - 24

    8 - H - 8

    c) H2 + I2 ↔ 2HI

    2 - I - 2

    2 - H - 2

    d) 4Fe + 3O2 ↔ 2Fe2O3

    4 - Fe - 4

    6 - O - 6

    e) 2Na + 2H2O ↔ 2NaOH + H2

    2 - Na - 2

    2 - O - 2

    4 - H - 4

    f) (NH4) 2Cr2O7 ↔ Cr2O3 + N2 + 4H2O

    2 - N - 2

    2 - Cr - 2

    7 - O - 7

    8 - H - 8

    g) P4 + 6Cl2 ↔ 4PCl3

    4 - P - 4

    12 - Cl - 12

    h) PtCl4 ↔ Pt + 2Cl2

    1 - Pt - 1

    4 - CL - 4
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