When 0.784 g of sodium metal is added to an excess of hydrochloric acid, 8150 J...

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Question

Answer 1

Answer #1

2 Na (s) + 2HCl (aq) --------> 2 NaCl (aq) + H₂ (g)

0.784 grams of sodium on reaction with excess of HCl liberates 8150 Joules of energy.

2*23 grams of sodium on reaction with excess of HCl liberates how many Joules of energy?

Amount of energy released

= 2*23 grams *8150 J/0.784 grams

= 477188.8 Joules

= 477.1888 kJ

= 477.2 kJ

Enthalpy of given reaction = 477.2 kJ or 238.6 kJ/mol

Answer 2

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