You have 75.0ml of a 2.50m solution

The reaction of Na2CrO4 (aq) and AgNO3 (aq) is as follows:
2AgNO3 (aq) +Na2CrO4 (aq) ⇒Ag2CrO4 (s) + 2NaNO3 (aq).

During this reaction, part of the CrO42- reacts with Ag+ and precipitates out of the solution, part of the CrO42- (excess amount) remains in the solution.

To find out how much CrO42- is reacted:
Moles of initial CrO42- = 0.075 L * 2.5 M = 0.1875 mole
Moles of initial Ag+ = 0.125 L * 1.79 M = 0.2238 mole

The reaction ratio between CrO42- and Ag+ is 1:2 according to the equation. So moles of CrO42- that is reacted is 0.2238 mole/2 = 0.1119 mole. Therefore, moles of CrO42- that remains in the solution is 0.1875mole-0.1119mole = 0.0756 mole

So the final concentration of CrO42- in the solution is 0.0756mole/(0.075L+0.125L)= 0.378 M.

Oxygen= 88ghydrogen= 12g