Answer:
7.26 moles of NH₃ are formed in this reaction
Explanation:
This is about the reaction for the production of ammonia
1 mol of nitrogen gas reacts to 3 moles of hydrogen in order to produce 2 moles of ammonia.
The equation is: N₂ + 3H₂ → 2NH₃
In the question, we were informed that the excess is the H₂ so the N₂ is limiting reagent. We determine the moles, that has reacted:
101.7 g / 28 g/mol = 3.63 moles
So, If 1 mol of nitrogen gas can produce 2 moles of ammonia
3.63 moles of N₂ must produce ( 2 . 3.63) / 1 = 7.26 moles of NH₃
Answer:
In this reaction, 7.26 moles of NH3 will be formed.
Explanation:
Step 1: Data given
Mass of N2 = 101.7 grams
Molar mass N2 = 28.0 g/mol
H2 is in excess
Molar mass H2 = 2.02 g/mol
Molar mass of NH3 = 17.03 g/mol
Step 2: The balanced equation
N2(g) + 3H2(g) → 2NH3(g)
Step 3: Calculate moles N2
Moles N2 = mass N2/ molar mass N2
Moles N2 = 101.7 grams / 28.0 g/mol
Moles N2 = 3.63 moles
Step 4: Calculate moles NH3
For 1 mol N2 we need 3 moles H2 to produce 2 moles NH3
For 3.63 moles N2 we'll produce2*3.63 = 7.26 moles NH3
In this reaction, 7.26 moles of NH3 will be formed.
