The detailed version of the question:
If we burn a strip of magnesium in air, there will be two main products, magnesium oxide and magnesium nitride. If we then treat that mixture with water, the nitride ion is converted to ammonia. After boiling away the water and ammonia, we would be left with only magnesium oxide. Knowing that air is approximately 78% nitrogen and 20% oxygen, we might expect that the mixture that forms in the burning process would be a similar distribution of magnesium nitride and magnesium oxide. In fact, the ratio is closer to 75% of the nitride to 25% of the oxide. Assuming that the reactions are 100% efficient (i.e., no loss of magnesium), how much magnesium oxide was formed initially, if the final product contains 3.22 grams of magnesium oxide?
After thinking about this for a while and working it out, it seems like the other information provided was irrelevant to finding the amount of MgO. I believe since the reactions are 100% efficient, that the final amount of 3.22 grams of MgO would be exactly the same for the initial amount of MgO, which was left over at the part when ammonia and water was evaporated. However, I am not sure; I would like to know, do the given percentages of nitrogen and oxygen given affect the unknown amount of MgO?