When 0.617 g of sodium metal is added to an excess of hydrochloric acid, 6410 J of heat are produced. What is the enthalpy of the reaction as written?
2Na(s)+2HCl(aq)⟶2NaCl(aq)+H2(g)
Mole of sodium=mass÷molar mass
Mole= 0.617÷23=0.026
∆H=q÷n
∆H= 6410÷0.026=246538.46J
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