When 0.607 g of sodium metal is added to an excess of hydrochloric acid, 6310 J of heat are produced. What is the enthalpy of the reaction as written?

Question

When 0.607 g of sodium metal is added to an excess of hydrochloric acid, 6310 J of heat are produced. What is the enthalpy of the reaction as written?

2 Na(s)+2 HCl(aq)⟶2 NaCl(aq)+H₂(g)

Enthalpy of reaction: _________kJ

Answer 1

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

0.607 g Na x 1 mol Na / 22.99 g = 0.0246 moles Na

The amount of heat produced from 0.0246 mols Na is 6310 J, but the reaction as written has 2 moles of Na, so to find the heat produced from 2 moles of Na, we do the following...

6310 J / 0.0246 mols Na x 2 mols Na = 478,000 J = 478 kJ = enthalpy of reaction as written