Question

Solid sodium reacts with water to form sodium hydroxide and hydrogen gas, according to the following equation.

2 Na(s) + 2 H₂O(l) → 2 NaOH(aq) + H₂(g)

If the reaction is carried out with 1.12 grams of sodium and excess water at 1.27 atm and 0°C, determine the amount of pressure-volume work done (in joules) during the production of hydrogen gas.

Answer 1

Let us consider the given data,

Mass of sodium = 1.12 g

Pressure = 1.27 atm

Temperature = 0⁰C + 273

= 273 K

We know molar mass of sodium = 23 g / mol

Let us consider given reaction ,

2 Na + 2 H2O -----------> 2 NaOH + H2

moles of sodium = mass of sodium / molar mass of sodium

= 1.12 / 23

= 0.0486

From the reaction, 2 moles of Na gives 1 mol of H2

0.0486 moles of Na gives ?

The moles of H2 = 0.0486 / 2

= 0.0243 mol

Let us consider, PV = n x R x T

1.27 x V = 0.0243 x 0.0821 x 273

V = (0.0243 x 0.0821 x 273) / 1.27

Volume , V = 0.428 L

Work done = P x V

= 1.27 x 0.428

= 0.5435 L.atm

W = 54.3 J

The amount of work done = 54.3 J