2 mole of N2 is reacted with 4 moles of H2 to produce ammonia . Mass of ammonia produced is
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Answer:
The chemical reaction can be given as :
Nitrogen + Hydrogen ---> Ammonia
N2+ 3H2----> 2NH3
28g of nitrogen reacts with 6g of water to produce 36 grams of water
1 mole of N2 + 3 moles of hydrogen ---> 2 moles of water
moles of N2 = 14x1mole/28 =1/2 mole of N2
moles of hydrogen=4x1mole/2=2 moles of hydrogen
Based on mole ration, enough H2 is present , limiting reactant is N2 and excess reactant is H2
So,
28 g of N2----> 36 g of Water
14g of N2--------?
=14x36/28
=18 grams of water
Therefore . when 14g of ammonia reacts mass of water produced is 18g
Explanation:
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Correct option is
Correct option isD
Correct option isD2.463 atm
Correct option isD2.463 atmN
Correct option isD2.463 atmN 2
Correct option isD2.463 atmN 2
Correct option isD2.463 atmN 2 +3H
Correct option isD2.463 atmN 2 +3H 2
Correct option isD2.463 atmN 2 +3H 2
Correct option isD2.463 atmN 2 +3H 2 →2NH
Correct option isD2.463 atmN 2 +3H 2 →2NH 3
Correct option isD2.463 atmN 2 +3H 2 →2NH 3
Correct option isD2.463 atmN 2 +3H 2 →2NH 3
Correct option isD2.463 atmN 2 +3H 2 →2NH 3 1 4 0 Before reaction
Correct option isD2.463 atmN 2 +3H 2 →2NH 3 1 4 0 Before reactionN
Correct option isD2.463 atmN 2 +3H 2 →2NH 3 1 4 0 Before reactionN 2
Correct option isD2.463 atmN 2 +3H 2 →2NH 3 1 4 0 Before reactionN 2
Correct option isD2.463 atmN 2 +3H 2 →2NH 3 1 4 0 Before reactionN 2 −H
Correct option isD2.463 atmN 2 +3H 2 →2NH 3 1 4 0 Before reactionN 2 −H 2
Correct option isD2.463 atmN 2 +3H 2 →2NH 3 1 4 0 Before reactionN 2 −H 2
Correct option isD2.463 atmN 2 +3H 2 →2NH 3 1 4 0 Before reactionN 2 −H 2 −NH
Correct option isD2.463 atmN 2 +3H 2 →2NH 3 1 4 0 Before reactionN 2 −H 2 −NH 3
Correct option isD2.463 atmN 2 +3H 2 →2NH 3 1 4 0 Before reactionN 2 −H 2 −NH 3
Correct option isD2.463 atmN 2 +3H 2 →2NH 3 1 4 0 Before reactionN 2 −H 2 −NH 3
Correct option isD2.463 atmN 2 +3H 2 →2NH 3 1 4 0 Before reactionN 2 −H 2 −NH 3 0 1 2 After reaction
Correct option isD2.463 atmN 2 +3H 2 →2NH 3 1 4 0 Before reactionN 2 −H 2 −NH 3 0 1 2 After reactionSince all the ammonia form get dissolved in water, no of moles of gas left =1
Correct option isD2.463 atmN 2 +3H 2 →2NH 3 1 4 0 Before reactionN 2 −H 2 −NH 3 0 1 2 After reactionSince all the ammonia form get dissolved in water, no of moles of gas left =1P=
Correct option isD2.463 atmN 2 +3H 2 →2NH 3 1 4 0 Before reactionN 2 −H 2 −NH 3 0 1 2 After reactionSince all the ammonia form get dissolved in water, no of moles of gas left =1P= 10
Correct option isD2.463 atmN 2 +3H 2 →2NH 3 1 4 0 Before reactionN 2 −H 2 −NH 3 0 1 2 After reactionSince all the ammonia form get dissolved in water, no of moles of gas left =1P= 101×0.0821×300
Correct option isD2.463 atmN 2 +3H 2 →2NH 3 1 4 0 Before reactionN 2 −H 2 −NH 3 0 1 2 After reactionSince all the ammonia form get dissolved in water, no of moles of gas left =1P= 101×0.0821×300
Correct option isD2.463 atmN 2 +3H 2 →2NH 3 1 4 0 Before reactionN 2 −H 2 −NH 3 0 1 2 After reactionSince all the ammonia form get dissolved in water, no of moles of gas left =1P= 101×0.0821×300 =2.463 atm
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