The number of terminal oxygen atoms present in the product B obtained from the following reaction is ______.
FeCr$$_2$$O$$_4$$ + Na$$_2$$CO$$_3$$ + O$$_2$$ $$\rightarrow$$ A + Fe$$_2$$O$$_3$$ + CO$$_2$$
A + H$$^+$$ $$\rightarrow$$ B + H$$_2$$O + Na$$^+$$

Step 1: Identify product A from the first reaction.

The chromite ore reaction with sodium carbonate and oxygen (fusion):

$$4\text{FeCr}_2\text{O}_4 + 8\text{Na}_2\text{CO}_3 + 7\text{O}_2 \rightarrow 8\text{Na}_2\text{CrO}_4 + 2\text{Fe}_2\text{O}_3 + 8\text{CO}_2$$

So product A is sodium chromate ($$\text{Na}_2\text{CrO}_4$$).

Step 2: Identify product B from the second reaction.

When sodium chromate reacts with acid:

$$2\text{Na}_2\text{CrO}_4 + 2\text{H}^+ \rightarrow \text{Na}_2\text{Cr}_2\text{O}_7 + \text{H}_2\text{O} + 2\text{Na}^+$$

So product B is sodium dichromate ($$\text{Na}_2\text{Cr}_2\text{O}_7$$).

Step 3: Count terminal oxygen atoms in $$\text{Cr}_2\text{O}_7^{2-}$$.

In the dichromate ion, there are 7 oxygen atoms total. One oxygen atom bridges the two chromium atoms, while the remaining 6 oxygen atoms are terminal (3 on each Cr).

Therefore, the number of terminal oxygen atoms in product B is 6.