When molten Na hydroxide was electrolysed between inert electrodes, a gas was produced as a product. Identify the substance and gas produced, write the ionic equation at the cathode, and calculate the no of faraday used, no of mole of Na atoms, and volume of gas produced.

Step 1: Analyze the electrolysis of molten NaH.
Molten sodium hydride ($\text{NaH}$) dissociates into $\text{Na}^+$ ions and $\text{H}^-$ ions.
At the cathode (negative electrode), reduction occurs: $\text{Na}^+ (l) + e^- \to \text{Na} (l)$
At the anode (positive electrode), oxidation occurs: $2\text{H}^- (l) \to \text{H}_2 (g) + 2e^-$

a) For inert electrodes, materials that do not react with the electrolyte or products are used.
The electrodes could be made of graphite or platinum.

b) The gas produced at the anode is hydrogen gas ($\text{H}_2$).

c) The ionic equation for the reaction occurring at the cathode is:

$$\text{Na}^+ (l) + e^- \to \text{Na} (l)$$

d) Given:
Current ($I$) = 1.5 A
Time ($t$) = 20 minutes
Faraday constant ($F$) = 96000 C/mol $e^-$
Molar volume of gas at rtp = 24 $\text{dm}^3/\text{mol}$

i) Calculate the number of Faradays used in the experiment.
First, convert time to seconds:

$$t = 20 \text{ min} \times 60 \text{ s/min} = 1200 \text{ s}$$

Next, calculate the total charge ($Q$) passed:

$$Q = I \times t = 1.5 \text{ A} \times 1200 \text{ s} = 1800 \text{ C}$$

Now, calculate the number of Faradays:

$$\text{Number of Faradays} = \frac{Q}{\text{Faraday constant}} = \frac{1800 \text{ C}}{96000 \text{ C/F}} = 0.01875 \text{ F}$$

The number of Faradays used is $\boxed{\text{0.01875 F}}$.

ii) Calculate the number of moles of Na atom produced.
From the cathode reaction: $\text{Na}^+ (l) + e^- \to \text{Na} (l)$.
1 mole of $\text{Na}$ is produced by 1 mole of electrons (1 Faraday).

$$\text{Moles of Na} = \text{Number of Faradays} = 0.01875 \text{ mol}$$

The number of moles of Na atom produced is $\boxed{\text{0.01875 mol}}$.

iii) Calculate the volume of gas produced at rtp.
From the anode reaction: $2\text{H}^- (l) \to \text{H}_2 (g) + 2e^-$.
1 mole of $\text{H}_2$ gas is produced by 2 moles of electrons (2 Faradays).

$$\text{Moles of H}_2 = \frac{\text{Number of Faradays}}{2} = \frac{0.01875 \text{ F}}{2} = 0.009375 \text{ mol}$$

At rtp, 1 mole of gas occupies 24 $\text{dm}^3$.

$$\text{Volume of H}_2 = \text{Moles of H}_2 \times \text{Molar volume at rtp}$$ $$\text{Volume of H}_2 = 0.009375 \text{ mol} \times 24 \text{ dm}^3/\text{mol} = 0.225 \text{ dm}^3$$

The volume of gas produced at rtp is $\boxed{\text{0.225 dm}^3}$.