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Consider the reaction 2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g) When 2 moles of Na react with water at 25°C and 1 atm, the volume of H2 formed is 24.5 L. Calculate the magnitude of work done in joules when 0.55 g of Na reacts with water under the same conditions. (The conversion factor is 1 L · atm = 101.3 J.)

Respuesta :

Answer : The value of work done for the system is, 29.7 J

Explanation :

First we have to calculate the moles of [tex]Na[/tex]

[tex]\text{Moles of }Na=\frac{\text{Mass of }Na}{\text{Molar mass of }NaN_3}[/tex]

Molar mass of [tex]Na[/tex] = 23 g/mole

[tex]\text{Moles of }Na=\frac{0.55g}{23g/mole}=0.0239mole[/tex]

Now we have to calculate the volume of hydrogen.

As, 2 mole of Na produced 24.5 L volume of hydrogen gas

So, 0.0239 mole of Na produced [tex]\frac{0.0239}{2}\times 24.5=0.293L[/tex] volume of hydrogen gas

Now we have to calculate the magnitude of work done.

Work done = Pressure × Volume

Work done = 1.00 atm × 0.293 L

Work done = 0.293 L.atm

Conversion used : (1 L.atm = 101.3 J)

Work done = 0.293 × 101.3 = 29.7 J

Therefore, the value of work done for the system is, 29.7 J

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