Question

7 la production du sel de table pourrait se faire selon l’équation thermochimique suivante. 2 na$_{(s)}$ + cl$_{2(g)}$ → 2 nacl$_{(s)}$ δh = -411,2 kj/mol de nacl on veut produire 350 g de sel de table. quelle quantité d’énergie sera produite ?

Explanation:

Step1: Calculate molar mass of NaCl

Molar mass of Na: \(22.99\space g/mol\), Cl: \(35.45\space g/mol\).
Molar mass of \(NaCl = 22.99 + 35.45 = 58.44\space g/mol\).

Step2: Find moles of NaCl

Given mass of NaCl: \(350\space g\).
Moles of \(NaCl = \frac{350\space g}{58.44\space g/mol} \approx 5.99\space mol\) (≈6 mol).

Step3: Relate moles to energy

From reaction: \(2\space mol\) NaCl releases \(4111.2\space kJ\) (since \(\Delta H = -4111.2\space kJ/mol\) of NaCl? Wait, wait, the reaction is \(2Na + Cl_2
ightarrow 2NaCl\), \(\Delta H = -4111.2\space kJ/mol\) of \(2\space mol\) NaCl? Wait, the problem says \(\Delta H = -4111.2\space kJ/mol\) of NaCl? Wait, the equation: \(2\space mol\) NaCl is produced, \(\Delta H = -4111.2\space kJ\) (so per 2 mol NaCl, energy is -4111.2 kJ, meaning 4111.2 kJ released for 2 mol NaCl).

So for 2 mol NaCl: energy released = 4111.2 kJ.
For \(n\) mol NaCl, energy released \(E = \frac{4111.2\space kJ}{2\space mol} \times n\space mol\).

We have \(n \approx 5.99\space mol\) (≈6 mol). Wait, 350g NaCl: \(350 / 58.44 ≈ 5.99\) mol ≈6 mol.

So \(E = \frac{4111.2}{2} \times 5.99 ≈ 2055.6 \times 5.99 ≈ 12313\space kJ\) (approx). Wait, let's do exact:

Moles of NaCl: \(350 / 58.44 ≈ 5.989\space mol\).

Energy per 2 mol NaCl: 4111.2 kJ. So per 1 mol NaCl: \(4111.2 / 2 = 2055.6\space kJ/mol\).

Thus, energy released: \(2055.6\space kJ/mol \times 5.989\space mol ≈ 2055.6 \times 6 ≈ 12333.6\space kJ\) (approx 12300 kJ, or more accurately: \(2055.6 \times 5.989 ≈ 12310\space kJ\)).

Answer:

Approximately \(12300\space kJ\) (or more precisely, using exact moles: \( \frac{4111.2}{2} \times \frac{350}{58.44} ≈ \frac{4111.2 \times 350}{2 \times 58.44} ≈ \frac{1,438,920}{116.88} ≈ 12310\space kJ \)) of energy is produced (released).