Question
Question: Calculate the volume of ammonia gas produced at STP when 140 g of nitrogen gas reacts with 30g hydro...
Calculate the volume of ammonia gas produced at STP when 140 g of nitrogen gas reacts with 30g hydrogen gas. (atomic mass: N =14 u, H= 1u)
Solution
For determining the volume of ammonia gas:
-First find the number of moles of N2 and H2.
-Then find the number of moles after the reaction.
-Use the formula: volume of ammonia gas = number of moles × volume at STP.
Complete step by step solution:
We have been provided that nitrogen gas has reacts with hydrogen gas to form ammonia gas:
N2+3H2→2NH3
So, for finding the volume of ammonia gas,
Number of moles of N2 in 140g: 28140=5moles (molar mass of N is 14g/mole)
Number of moles of H2 in 30g: 230=15moles (molar mass of H is 1g/mole)
Now, at time=0, moles would be:
For the reaction: N2+3H2→2NH3,
Moles of N2and H2 are: 5 and 15 respectively,
So, after the reaction, moles would be:
Moles of N2: 5−(5×1),
Moles of H2: 15−(3×5),
Moles of NH3:2×5,
So, the moles of ammonia come out to be: 10,
Now, we have been provided standard conditions for the formation of ammonia gas that is STP,
Standard temperature and pressure are standard sets of conditions for experimental measurements to be established to allow comparisons to be made between different sets of data.
Now we will be finding the volume of ammonia gas,
So, for that, we will be using the formula: volume of ammonia gas = number of moles × volume at STP. (volume at STP= 22.4L)
So, the volume would be: volume=10×22.4L=224L,
So, the volume of ammonia gas comes out to be: 224L.
Note: Ammonia is industrially prepared by the Haber process, a chemical method that uses nitrogen gas and hydrogen gas to synthesize ammonia. One nitrogen gas molecule reacts with three hydrogen gas molecules over finely divided iron as a catalyst to produce two ammonia molecules