Sodium reacts with water to give hydrogen gas and sodium hydroxide solution - HSC - SSCE Chemistry - Question 14 - 2013 - Paper 1

The balanced chemical equation for the reaction of sodium with water is:

$\text{2 Na (s) + 2 H}_2\text{O (l) \rightarrow 2 NaOH (aq) + H}_2 \text{(g)}$

From the equation, 2 moles of sodium produce 1 mole of hydrogen gas. Therefore, 1 mole of sodium will produce:

$\text{Volume of } H_2 = \frac{1 \, \text{mol Na}}{2} = 0.5 \, \text{mol } H_2$

To find the volume of hydrogen gas at 25°C (298 K) and 100 kPa, we use the ideal gas law:

$PV = nRT$

Where:

• $P$ = pressure in kPa = 100 kPa
• $V$ = volume in L (to be calculated)
• $n$ = number of moles of $H_2$ = 0.5 mol
• $R$ = ideal gas constant ≈ 8.314 J/(mol·K) = 8.314 L·kPa/(mol·K)
• $T$ = temperature in K = 25°C + 273 = 298 K.

Rearranging the ideal gas equation gives:

$V = \frac{nRT}{P} = \frac{0.5 \, \text{mol} \times 8.314 \, \text{L·kPa/(mol·K)} \times 298 \, \text{K}}{100 \, \text{kPa}}$

Calculating this yields: $V = \frac{1239.74}{100} = 12.40 \, \text{L}$

Based on the calculation, the volume of gas produced from the reaction is 12.40 L, which corresponds to option (B).