A sample of sodium dichromate, Na2Cr2O7, is placed into a container by itself. The sample of material in the container is analyzed, and it is found to contain exactly 0.67 moles of sodium atoms. How many moles of oxygen atoms are in this sample?

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