Question

when 0.503 g of sodium metal is added to an excess of hydrochloric acid, 5230 j of heat are produced. what is the enthalpy of the reaction as written?
2 na(s) + 2 hcl(aq) → 2nacl(aq) + h₂(g)

Explanation:

Step1: Calculate the moles of Na

The molar mass of Na is \(M = 22.99\space g/mol\). Using the formula \(n=\frac{m}{M}\), where \(m = 0.503\space g\).
\(n=\frac{0.503\space g}{22.99\space g/mol}\approx0.0219\space mol\)

Step2: Relate moles of Na to the reaction

From the balanced equation \(2Na(s)+2HCl(aq)\to2NaCl(aq)+H_2(g)\), when \(n = 2\space mol\) of Na react.
We know that for \(0.0219\space mol\) of Na, heat produced \(q = 5230\space J\). Let \(\Delta H\) be the enthalpy for \(2\space mol\) of Na.
Using the proportion \(\frac{\Delta H}{2\space mol}=\frac{5230\space J}{0.0219\space mol}\)
\(\Delta H=\frac{5230\space J\times2\space mol}{0.0219\space mol}\)
\(\Delta H=- 478000\space J/mol=-478\space kJ/mol\) (negative because heat is produced, exothermic reaction)

Answer:

\(-478\space kJ/mol\)