Question

la production de sel de table pourrait se faire selon léquation thermochimique suivante :
2 na + cl₂ → 2 nacl δh = -411,2 kj/mol de nacl
on produit 390 g de sel de table. quelle quantité dénergie sera produite ?

Explanation:

Step1: Find 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: Calculate moles of NaCl

Given mass of NaCl = \(390\space g\).
Moles \(n=\frac{mass}{molar\space mass}=\frac{390}{58.44}\approx6.67\space mol\).

Step3: Relate moles to energy from reaction

Reaction: \(2Na + Cl_2 ightarrow2NaCl\), \(\Delta H = -411.2\space kJ/mol\) (per 2 mol NaCl? Wait, recheck: The \(\Delta H\) is given as \(-411.2\space kJ/mol\) of NaCl? Wait, the reaction produces 2 mol NaCl with \(\Delta H = -411.2\space kJ\)? Wait, the problem says "\(\Delta H = -411.2\space kJ/mol\) de NaCl" (per mole of NaCl). Wait, let's confirm:

Wait, the reaction is \(2Na + Cl_2 ightarrow2NaCl\), so 2 moles of NaCl are produced. But the \(\Delta H\) is given as \(-411.2\space kJ\) per mole of NaCl? Wait, maybe the \(\Delta H\) is for 2 moles? Wait, no, the text says "\(\Delta H = -411.2\space kJ/mol\) de NaCl" – so per mole of NaCl.

So for 1 mole of NaCl, energy released is \(411.2\space kJ\) (since \(\Delta H\) is negative, exothermic).

So for \(6.67\space mol\) of NaCl, energy released \(E = n\times|\Delta H|\) (since \(\Delta H\) is per mole). Wait, no: Wait, the reaction's \(\Delta H\) is \(-411.2\space kJ\) when 1 mole of NaCl is formed? Wait, let's re-express:

From the reaction, 2 moles of NaCl are formed. But the problem states \(\Delta H = -411.2\space kJ/mol\) de NaCl – so per mole of NaCl. So for each mole of NaCl produced, 411.2 kJ is released (since \(\Delta H\) is negative, exothermic).

So moles of NaCl = \(6.67\space mol\).
Energy released \(E = 6.67\space mol\times411.2\space kJ/mol\approx2743\space kJ\)? Wait, no, wait: Wait, maybe the \(\Delta H\) is for 2 moles of NaCl. Let's check the units again. The problem says "\(\Delta H = -411.2\space kJ/mol\) de NaCl" – so per mole of NaCl. So if 2 moles of NaCl are produced, the total \(\Delta H\) would be \(2\times(-411.2)\space kJ\)? No, that can't be. Wait, maybe the \(\Delta H\) is \(-411.2\space kJ\) for 2 moles of NaCl? Let's see:

Wait, the reaction is \(2Na + Cl_2 ightarrow2NaCl\), so 2 moles of NaCl. If \(\Delta H = -411.2\space kJ\) for 2 moles, then per mole it's \(-205.6\space kJ/mol\). But the problem says "\(\Delta H = -411.2\space kJ/mol\) de NaCl" – so per mole. So we'll go with the problem's statement: \(\Delta H = -411.2\space kJ\) per mole of NaCl (exothermic, so energy released is positive).

So moles of NaCl: \(n = 390\space g / 58.44\space g/mol \approx 6.67\space mol\).

Energy released: \(E = 6.67\space mol \times 411.2\space kJ/mol \approx 2743\space kJ\). Wait, but let's check with 2 moles: If the reaction produces 2 moles of NaCl, and \(\Delta H = -411.2\space kJ\) (for 2 moles), then per mole it's \(-205.6\space kJ/mol\). But the problem says "\(\Delta H = -411.2\space kJ/mol\) de NaCl" – so maybe the problem has a typo, but we'll follow the given data.

Wait, let's recalculate:

Molar mass of NaCl: \(58.44\space g/mol\).
Moles of NaCl: \(390 / 58.44 \approx 6.67\space mol\).
Energy per mole of NaCl: \(411.2\space kJ\) (since \(\Delta H\) is -411.2, so energy released is 411.2 kJ per mole).
Total energy: \(6.67 \times 411.2 \approx 2743\space kJ\). Wait, but maybe the \(\Delta H\) is for 2 moles. Let's check the reaction: \(2Na + Cl_2 ightarrow2NaCl\), so 2 moles of NaCl. If \(\Delta H = -411.2\space kJ\) for 2 moles, then per mole it's \(-205.6\space kJ/mol\). But the problem says "\(\Delta H = -411.2\space kJ/mol\) de NaCl" – so per mole. So we'll proceed with that.

Alternatively, maybe the \(\Delta H\) is -411.2 kJ for 2 moles of NaCl. Let's check:

If 2 moles of NaCl have \(\Delta H = -411.2\space kJ\), then per mole it's \(-205.6\space kJ/mol\). Then for 6.67 moles (which is 390g / 58.44g/mol ≈6.67), the energy would be (6.67 / 2) * 411.2 ≈ (3.335) * 411.2 ≈1371 kJ. But this contradicts the problem's \(\Delta H\) unit.

Wait, the original problem in French: "La production de sel de table pourrait se faire selon l'équation thermochimique suivante: \(2Na + Cl_2 ightarrow2NaCl\), \(\Delta H = -411,2\space kJ/mol\) de NaCl. On produit 390g de sel de table. Quelle quantité d'énergie sera produite?"

So \(\Delta H\) is -411.2 kJ per mole of NaCl. So when 1 mole of NaCl is formed, 411.2 kJ is released (exothermic).

So moles of NaCl: 390g / 58.44g/mol ≈6.67 mol.

Energy released: 6.67 mol * 411.2 kJ/mol ≈2743 kJ.

Wait, but let's check the calculation again:

390 divided by 58.44: 58.44*6=350.64, 390-350.64=39.36, 39.36/58.44≈0.67, so total≈6.67 mol.

6.67 * 411.2 ≈6.67*400=2668, 6.67*11.2≈74.7, total≈2668+74.7≈2742.7 kJ≈2740 kJ or 2.74×10³ kJ.

But maybe the \(\Delta H\) is for 2 moles. Let's see: If 2 moles of NaCl have \(\Delta H = -411.2\space kJ\), then per mole it's -205.6 kJ/mol. Then 6.67 mol would be (6.67/2)*411.2≈3.335*411.2≈1371 kJ. But the problem says "per mole of NaCl", so we'll go with the first calculation.

Answer:

Approximately \(\boldsymbol{2740\space kJ}\) (or more precisely, \(2743\space kJ\)) of energy is produced.