1
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ This seems like it might be a homework question or have come from a textbook or similar source. Check out our [homework policy] so you can improve this question. Primarily, what work have you done and where are you stuck? $\endgroup$
    – Ben Norris
    Jan 15, 2014 at 21:17

2 Answers 2

1
$\begingroup$

Maybe I'm just too tired right now, but didn't you mention that the absolute amount of magnesium oxide doesn't change during aqueous workup?

Isn't thus the amount of 3.22 g $\ce{MgO}$ exactly what was formed initially?

Wouldn't it be more reasonable to let the students calculate how much magnesium metal was burned initially, taking the $\ce{N2}$/$\ce{O2}$ ratio in air and the stoichiometry of the magnesium nitride into account? ;)

$\endgroup$
1
  • 1
    $\begingroup$ Hm, it just might be that my professor was adding extra information for distraction. I had thought the same with the final amount of 3.22 g of MgO being the same as the initial, but I was not sure with all of the other information provided. $\endgroup$ Jan 15, 2014 at 21:00
0
$\begingroup$

The magnesium nitride hydrolyzes to Mg(OH)2 then dehydrates to MgO. If formation is 3:1, then 4x = 3.22 g MgO. Solve for x. All the fun is in the footnotes.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.