Question

when 0.815 g of sodium metal is added to an excess of hydrochloric acid, 8470 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

Explanation:

Step1: Calculate the number of moles of sodium

The molar mass of \(Na\) is \(M = 22.99\space g/mol\).
The number of moles \(n=\frac{m}{M}\), where \(m = 0.815\space g\).
\(n=\frac{0.815\space g}{22.99\space g/mol}\approx0.03545\space mol\)

Step2: Relate the heat produced to the moles of sodium

From the reaction \(2Na(s)+2HCl(aq)\to2NaCl(aq) + H_{2}(g)\), when \(2\space mol\) of \(Na\) reacts, the enthalpy change is \(\Delta H\).
We know that for \(n = 0.03545\space mol\) of \(Na\), the heat produced \(q=8470\space J=8.47\space kJ\)
If \(2\space mol\) of \(Na\) reacts, let the enthalpy change be \(\Delta H\). Using the proportion \(\frac{\Delta H}{2\space mol}=\frac{q}{n}\)

Step3: Solve for \(\Delta H\)

Substitute \(n = 0.03545\space mol\) and \(q = 8.47\space kJ\) into \(\Delta H=\frac{2\space mol\times q}{n}\)
\(\Delta H=\frac{2\times8.47\space kJ}{0.03545\space mol}\approx - 479\space kJ\) (negative because heat is produced, exothermic reaction)

Answer:

\(-479\space kJ\)