form 4 chemistry ex plz help

2009-11-30 6:36 am
when magnesium power is heated strongly in a current of nitrogen , a yellow solid magnesium nitride(Mg3N2) is formed. on treatment with water, this solid gives ammonia and magmesium oxide.
(a) write an equation for the formation of magnesium nitride.
(b) write an equation for the action of water on magnesium nitride.
(c) find the mass of magnesium nitride, which could be obtained, from 10g of magnesium.
更新1:

(d) if the amount of magnesium nitride obtained in (c) is boiled with excess water, find(1) mass of magnesium oxide, (2) volume of ammonia produced

回答 (1)

2009-11-30 9:22 am
✔ 最佳答案
(a)
3Mg(s) + N2(g) → Mg3N2(s)


(b)
Mg3N2(s) + 3H2­O(l) → 3MgO(s) + 2NH3(g)


(c)
3Mg(s) + N2(g) → Mg3N2(s)
Mole ratio Mg : Mg3N2 = 3 : 1

Molar mass of Mg = 24.3 g/mol
No. of moles of Mg used = 10/24.3 = 0.4115 mol
No. of moles of Mg3N2 formed = 0.4115 x (1/3) = 0.1372 mol
Molar mass of Mg3N2 = 24.3x3 + 14x2 = 100.9 g/mol
Mass of Mg3N2 = 0.1372 x 100.9 = 13.8 g


(d)
(1)
Mg3N2(s) + 3H2­O(l) → 3MgO(s) + 2NH3(g)
Mole ratio Mg3N2 : MgO = 1 : 3

No. of moles of Mg3N2 used = 0.1372 mol
No. of moles of MgO formed = 0.1372 x 3 = 0.4116 mol
Molar mass of MgO = 24.3 + 16 = 40.3 g/mol
Mass of MgO formed = 0.4116 x 40.3 = 16.6 g

(2)
Mole ratio Mg3N2 : NH3 = 1 : 2

No. of moles of Mg3N2 used = 0.1372 mol
No. of moles of NH3 formed = 0.1372 x 2 = 0.2744 mol
Molar volume of NH3 in room conditions = 24 dm^3
Volume of NH3 formed = 0.2744 x 24 = 6.59 dm^3


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