The number of non-ionisable protons present in the product B obtained from the following reaction is ______
$$C_2H_5OH + PCl_3 \rightarrow C_2H_5Cl + A$$
$$A + PCl_3 \rightarrow B$$

We need to find the number of non-ionisable protons in product B.

When ethanol reacts with phosphorus trichloride, $$3C_2H_5OH + PCl_3 \rightarrow 3C_2H_5Cl + H_3PO_3$$. Therefore, product A is phosphorous acid ($$H_3PO_3$$).

Since phosphorous acid exists in its tautomeric form $$HP(O)(OH)_2$$, it contains:

- One $$P=O$$ double bond

- Two $$P-OH$$ bonds (ionisable protons)

- One $$P-H$$ bond (non-ionisable proton)

Subsequently, when $$H_3PO_3$$ reacts with $$PCl_3$$, the hydroxyl groups undergo condensation with chloride atoms, releasing HCl and forming a P-O-P linkage. The product is pyrophosphorous acid ($$H_4P_2O_5$$).

Pyrophosphorous acid consists of two phosphorus atoms joined by a bridging oxygen (P-O-P). Each phosphorus atom has:

- One $$P=O$$ bond

- One $$P-OH$$ bond (ionisable proton)

- One $$P-H$$ bond (non-ionisable proton)

- Shared bridging oxygen connecting to the other P

The molecular formula $$H_4P_2O_5$$ accounts for 2 OH protons (ionisable) and 2 P-H protons (non-ionisable), giving 4 total hydrogen atoms.

Therefore, the number of non-ionisable protons in product B is $$\boxed{2}$$.