2Mg + O2 ==> 2MgO
3Mg + N2 ==> Mg3N2
Mg3N2 + 6HOH ==> 3Mg(OH)2 + 2NH3
I would make up some convenient numbers (such as 24.3g Mg) and calculate how much MgO was produced and how much O in the MgO, then calculate the formula from that. Then go through the same calculation but change the O and see how that affects the formula.
Assume that some of the magnesium nitride is not completely converted to
magnesium hydroxide and subsequently to magnesium oxide. Will your empirical
formula be too high in magnesium or too high in oxygen? Please show any
calculations you used to determine your answer.
Where would you even start to answer this?
2 answers
How would you calculate how much MgO was produced? Would you just use the first reaction 2Mg+O2 --> 2MgO or would you have to put the amount of Mg through the other reactions with the mole to mole ratios and then find the amount of MgO produced?
1 answer