The following equation represents the reaction between ammonia and chlorine:
8NH3(g) + 3Cl2(g) → 6NH4Cl(s) + N2(g)
15.3 g of ammonia and 21.3 g of chlorine are allowed to react.
One of the reaction is in excess.
Determine ....
(a) The limiting reactant ;
(b) The mass of nitrogen produced ;
(c) The mass of excess reactant left after the reaction.
Thankyou!! Orz
chem ....
2008-05-05 9:28 am
回答 (1)
2008-05-05 9:55 am
✔ 最佳答案
a) no.of moles of NH3 =15.3/ [14 +3] =0.9
no.of moles of Cl2 =21.3/ [35.5 x2]=0.3
Cl2 is the limiting reactant.
b)
8NH3(g) + 3Cl2(g) → 6NH4Cl(s) + N2(g)
3 moles of Cl2 = 1mole of N2
no.of moles of N2 =0.3 /3 =0.1
Mass of N2 produced =0.1 x 14(2)=2.8g
c)
The reactant left is NH3.
mass of excess reactant left = (0.9/8 -0.3/3)x17 =0.213g
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收錄日期: 2021-05-03 06:34:08
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