Answer:
The answer to your question is a) N₂ b) 3.04 g of NH₃
Explanation:
Data
mass of H₂ = 2.5 g
mass of N₂ = 2.5 g
molar mass H₂ = 2.02 g
molar mass of N₂ = 28.02 g
molar mass of NH₃ = 17.04 g
Balanced chemical reaction
3H₂ + 1 N₂ ⇒ 2NH₃
A)
Calculate the theoretical yield 3H₂ / N₂ = 3(2.02) / 28.02 = 0.22
Calculate the experimental yield H₂/N₂ = 2.5/2.5 = 1
Conclusion
The limiting reactant is N₂ (nitrogen) because the experimental proportion was higher than the theoretical proportion.
B)
28.02 g of N₂ -------------------- (2 x 17.04) g of NH₃
2.5 g of N₂ -------------------- x
x = (2.5 x 2 x 17.04) / 28.02
x = 85.2 / 28.02
x = 3.04 g of NH₃
1mol aluminium chloride gives 1mol aluminium and 3mol chloride
density equals mass divided by volume
d=m/v
m=v*d
=78.3*2.7
=211.41grams
HF and NaF - If the right concentrations of aqueous solutions are present, they can produce a buffer solution.
<h3>What are buffer solutions and how do they differ?</h3>
- The two main categories of buffers are acidic buffer solutions and alkaline buffer solutions.
- Acidic buffers are solutions that contain a weak acid and one of its salts and have a pH below 7.
- For instance, a buffer solution with a pH of roughly 4.75 is made of acetic acid and sodium acetate.
<h3>Describe buffer solution via an example.</h3>
- When a weak acid or a weak base is applied in modest amounts, buffer solutions withstand the pH shift.
- A buffer made of a weak acid and its salt is an example.
- It is a solution of acetic acid and sodium acetate CH3COOH + CH3COONa.
learn more about buffer solutions here
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Answer:
Ammonium nitrate, (NH4NO3), a salt of ammonia and nitric acid, used widely in fertilizers and explosives. The commercial grade contains about 33.5 percent nitrogen, all of which is in forms utilizable by plants; it is the most common nitrogenous component of artificial fertilizers.
