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7nadin3 [17]
3 years ago
6

Complete the balanced molecular chemical equation for the reaction below. If no reaction occurs, write NR after the reaction arr

ow. Pb ( N O ₃ ) ₂ (aq) + 2 K I (aq) → Pb I ₂ + 2 K N O ₃ ☐⁴⁻ ☐³⁻ ☐²⁻ ☐⁻ ☐⁺ ☐²⁺ ☐³⁺ ☐⁴⁺ 1 2 3 4 5 6 7 8 9 0 ☐₁ ☐₂ ☐₃ ☐₄ ☐₅ ☐₆ ☐₇ ☐₈ ☐₉ ☐₀ + ( ) → ⇌ (s) (l) (g) (aq) NR O K I N Pb Reset
Chemistry
1 answer:
Alex Ar [27]3 years ago
3 0

Explanation:

The reaction is given as;

Pb ( N O ₃ ) ₂ (aq) +  K I (aq) → Pb I ₂ +  K N O ₃

To ensure the reaction is balanced, there has to be equal number of toms of each element present in both the reactant and product side of the reaction.

This leads us to;

Pb ( N O ₃ ) ₂ (aq) + 2 K I (aq) → Pb I ₂(s) + 2 K N O ₃(aq)

Ths way all atoms of elements are the same.

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The enthalpy of reaction for the combustion of ethane 2CH₃CH₃ + 7O₂ → 4CO₂ + 6H₂O calculated from the average bond energies of the compounds is -2860 kJ/mol.

The reaction is:

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The enthalpy of reaction (1) is given by:

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The bonds of the compounds of reaction (1) are:

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Hence, the enthalpy of reaction (1) is (eq 2):

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\Delta H = 2*\Delta H_{CH_{3}CH_{3}} + 7\Delta H_{O_{2}} - (4*\Delta H_{CO_{2}} + 6*\Delta H_{H_{2}O})      

\Delta H = 2*(6*\Delta H_{C-H} + \Delta H_{C-C}) + 7\Delta H_{O=O} - (4*2*\Delta H_{C=O} + 6*2*\Delta H_{H-O})  

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\Delta H = -2860 kJ/mol          

Therefore, the enthalpy of reaction for the combustion of ethane is -2860 kJ/mol.

Read more here:

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