F.4 Chemistry Mole

[匿名]

2013-02-18 12:00 am

2.64g of ammonium sulphate were heated with excess calcium hydroxide and the folowing reaction occured:
(NH4)2SO4(s)+Ca(OH)2(s)→CaSO4(s)+2NH3(g)+2H2O(l)

The ammonia formed was passed over excess heated copper(II)oxide.
3CuO(s)+2NH3(g)→3Cu(s)+N2(g)+3H2O(l)

Assuming all the ammonia had reacted, calculate:
a) the mass of calcium hydroxide reacted with the ammonium sulphate
b)the mass of calcium sulphate formed
c)the mass of nitrogen gas produced

回答 (1)

賣女孩的火柴

2013-02-18 4:07 am

✔ 最佳答案

(a)
(NH4)2SO4 + Ca(OH)2 → CaSO4 + 2NH3+ 2H2O
No. of moles of (NH4)2SO4 reacted = 2.64/(14x2+ 1x8 + 32 + 16x4) = 0.02 mol
No. of moles of Ca(OH)2 reacted = 0.02 mol
Mass of Ca(OH)2 reacted = 0.02 x (40 + 16x2 + 1x2) = 1.48 g

(b)
(NH4)2SO4 + Ca(OH)2 → CaSO4 + 2NH3+ 2H2O
No. of moles of (NH4)2SO4 reacted = 0.02 mol
No. of moles of CaSO4 formed = 0.02 mol
Mass of CaSO4 formed = 0.02 x (40 + 32 + 16x4) = 2.72 g

(c)
(NH4)2SO4 + Ca(OH)2 → CaSO4 + 2NH3+ 2H2O
No. of moles of (NH4)2SO4 reacted = 0.02 mol
No. of moles of NH3 formed = 0.02 x 2 = 0.04 mol

3CuO + 2NH3 → 3Cu + N2 + 3H2O
No. of moles of NH3 reacted = 0.04 mol
No. of moles of N2 formed = 0.04 x (1/2) = 0.02 mol
Mass of N2 formed = 0.02 x (14x2) = 0.56 g

參考: 賣女孩的火柴

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