Mg_3 N_2(s)_(s)+3H_2 O_(l) 3MgO_(s)+2NH_(3(g) Trial 1 Trial 2 Mass of Mg in g ±0
ID: 542264 • Letter: M
Question
Mg_3 N_2(s)_(s)+3H_2 O_(l) 3MgO_(s)+2NH_(3(g)
Trial 1
Trial 2
Mass of Mg in g ±0.0001 g
0.3059g
0.2846g
Mass of crucible and lid + residue in g ±0.0001 g
28.9329g
29.6265g
Mass of residue in g ±0.0001 g
0.4788g
0.5016g
Mass of used oxygen in g ±0.0001 g
0.1729g
0.2170g
Moles of Mg
0.012586
0.011710
Moles of O
0.010806
0.013562
Normalized moles of Mg
1.1647 mol of Mg
1 mol of Mg
Normalized moles of O
1 mol of O
1.16 mol of O
Empirical formula
MgO
MgO
How do the theoretical and experimental empirical formula can be compare?
How an incomplete conversion of Mg3N2 affect the results
From the result, is this a plausible method to determine the formula of metal oxides?
Trial 1
Trial 2
Mass of Mg in g ±0.0001 g
0.3059g
0.2846g
Mass of crucible and lid + residue in g ±0.0001 g
28.9329g
29.6265g
Mass of residue in g ±0.0001 g
0.4788g
0.5016g
Mass of used oxygen in g ±0.0001 g
0.1729g
0.2170g
Moles of Mg
0.012586
0.011710
Moles of O
0.010806
0.013562
Normalized moles of Mg
1.1647 mol of Mg
1 mol of Mg
Normalized moles of O
1 mol of O
1.16 mol of O
Empirical formula
MgO
MgO
Explanation / Answer
.
1)The theoretical formula of MgO is that in which molar ratio of Mg and O is 1:1 but in experimental empirical formula the molar ratio of Mg and O is not exactly 1:1. It differs. For trial 1 the molar ratio of Mg and O is 1.1647:1 and for trial 2 the molar ratio of Mg and O is 1:1.16. For trial 1, the product is slightly magnesium-rich the ratio of Mg: O is greater than the 1-to-1 expected. For trial 2, the product is slightly magnesium-poor product would have a ratio of Mg: O that is less than the 1-to-1 expected.
2) Magnesium reacts vigorously when heated in the presence of air. Although there is a higher percentage of N2 gas in the atmosphere than O2 but oxygen is more reactive than N2. So, more magnesium oxide is formed. The small amount of nitride that forms can be removed with the addition of water, which converts the nitride to magnesium hydroxide and ammonia gas. Heating Mg(OH)2 causes the loss of water and conversion of the hydroxide to the oxide.
Mg3N2 + 3H2O --> 3MgO + 2NH3
Mg3N2(s) + H2O (l) ---> Mg(OH)2 + NH3(g)
Mg(OH)2 heat ----->MgO + H2O(l)
3) Yes, this is a plausible method to determine the formula of metal oxides