✔ 最佳答案
Molar mass of Mg = 24.3 g mol-1
Molar mass of N2 = 14x2 = 28 g mol-1
Molar mass of Mg3N2 = 24.3x3 + 14x2 = 100.9 g mol-1
3Mg + N2 → Mg3N2
Reacted mole ratio Mg : N2 = 3 : 1
No. of moles of Mg added = mass/(molar mass) = 4.86/24.3 = 0.2 mol
No. of moles of N2 added = mass/(molar mass) = 3.52/28 = 0.126 mol
Added mole ratio Mg : N2 = 0.2 : 0.126 = 3 : 1.875
Hence, N2 is in excess, and Mg is the limiting reactant (completely reacted).
3Mg + N2 → Mg3N2
Mole ratio Mg : Mg3N2 = 3 : 1
No. of moles of Mg reacted = 0.2 mol
No. of moles of Mg3N2 formed = 0.2 x (1/3) mol
Mass of Mg3N2 formed = mol x (molar mass) = 0.2 x (1/3) x 100.9 = 6.727 g
=
2008-12-04 11:24:39 補充：
A known amount of Mg and a known amount of N2 have been added. One of the two reactants is in excess, and another one is completely reacted (i.e. the limiting reactant). The limiting reactant is determined in part 3.
line1:
The amount of Mg added (4.86 g) to the reaction mixture is 0.2 mol.
2008-12-04 11:25:38 補充：
line 2:
The amount of N2 added (3.52 g) to the reaction mixture is 0.126 mol
line 3:
Moles of Mg added : Moles of N2 added = 0.2 : 0.126
Which one is in excess ?
2008-12-04 11:26:54 補充：
Refer to part 2, Moles of Mg reacted : Moles of N2 reacted = 3 : 1
This means that 1 mol of N2 is needed to completely react with 3 mol of Mg. In other words, 3 mol of Mg is needed to completely react with 1 mol of N2.
2008-12-04 11:29:22 補充：
Method 1 (as in my answer):
Moles of Mg added : Moles of N2 added = 0.2 : 0.126 = 3 : 1.875
Moles of Mg needed : Moles of N2 needed = 3 : 1
Compare the two ratio, N2 is in excess.
2008-12-04 11:31:17 補充：
Method 2:
Moles of Mg added : Moles of N2 added = 0.2 : 0.126 = 1.587 : 1
Moles of Mg needed : Moles of N2 needed = 3 : 1
Compare the two ratio. Mg is in sufficient (i.e. completely reacted)
2008-12-04 11:32:55 補充：
Method 3:
0.2 mol of Mg and 0.126 mol of N2 have been added.
If 0.2 mol of Mg is completely reacted,
No. of moles N2 needed = 0.2(1/3) = 0.0667 mol
Hence, N2 is in excess.
2008-12-04 11:33:40 補充：
Method 4:
0.2 mol of Mg and 0.126 mol of N2 have been added.
If 0.126 mol of N2 is completely reacted,
No. of moles Mg needed = 0.126(3) = 0.378 mol
Hence, Mg is insufficient (i.e. completely reacted).
