calculate the volume of ammonia gas produced at stp when 140 g of N gas reacted with 30 g of H gas?
Answers
Answered by
25
Moles of 140g N₂ = 140 / 28 = 5
Moles of 30g H₂ = 30 / 2 = 15
Balanced Equation for formation of Ammonia:
N₂ + 3H₂ ==> 2NH₃
From above equation we can say……,
5 moles of nitrogen and 15 moles of hydrogen when react will give 10 moles of Ammonia
Volume of 10 moles of Ammonia at STP
= 10 × 22.4 litres
= 224 litres
Volume of ammonia gas produced is 224 litres.
Moles of 30g H₂ = 30 / 2 = 15
Balanced Equation for formation of Ammonia:
N₂ + 3H₂ ==> 2NH₃
From above equation we can say……,
5 moles of nitrogen and 15 moles of hydrogen when react will give 10 moles of Ammonia
Volume of 10 moles of Ammonia at STP
= 10 × 22.4 litres
= 224 litres
Volume of ammonia gas produced is 224 litres.
Answered by
7
Answer:224
Explanation:
No of moles in 140g Nitrogen=140/28=5
No of moles of hydregon=30/2=15
N2+3H2>2NH3
1:3:2
5:15:10
Volume of10 Mole of Ammonia At STP=10*22.4=224 L
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