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 number of moles of Na

The molar mass of Na is \(M = 22.99\space g/mol\).
The number of moles \(n=\frac{m}{M}\), 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 equation

From the balanced equation \(2Na(s)+2HCl(aq)\to2NaCl(aq)+H_2(g)\), when \(2\space mol\) of \(Na\) react, the heat change is \(\Delta H\).
We know that for \(n = 0.0219\space mol\) of \(Na\), \(q=- 5230\space J\) (negative because heat is produced, exothermic reaction).

Let \(x\) be the heat change when \(2\space mol\) of \(Na\) react.
Using the proportion \(\frac{x}{2\space mol}=\frac{-5230\space J}{0.0219\space mol}\)

\(x=\frac{- 5230\space J\times2\space mol}{0.0219\space mol}\)

\(x\approx - 478000\space J/mol=-478\space kJ/mol\)

Answer:

\(\Delta H=-478\space kJ/mol\)