posted by Maggie .
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?
By its very nature this problem is confusing because the MgO can come from two sources; i.e., the nitride as well as the oxide. If I read it right, however, I would multiply it this way.
3.22 x 0.25 = 0.805 g initially from MgO
3.22 x 0.75 = 2.42 g initially Mg3N2 that formed MgO.l
Check: 0.805+2.42 = 3.22