Answer:
2.345 moles
Explanation:
From the stoichiometry of the formula, Na₂Cr₂O₇
It is clear that the 1 mole of sodium dichromate contains 2 moles of sodium, 2 moles of chromium and 7 moles of oxygen.
Thus, given that the compound contains 0.67 moles of sodium. Thus, applying unitary method as:
2 moles of sodium and 7 moles of oxygen are present in 1 mole of sodium dichromate.
Also,
1 mole of sodium and 7/2 moles of oxygen are present in 1/2 moles of sodium dichromate.
Thus,
0.67 moles of sodium and (7/2)*0.67 moles of oxygen are present in (1/2)*0.67 moles of sodium dichromate.
0.67 moles of sodium and 2.345 moles of oxygen are present in 0.335 moles of sodium dichromate.
So, moles of oxygen present = 2.345 moles
