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
H2SO4 + ZN ------- ZNSO4+ H2
(SO4)²The sulphate salt is formed......
Hope it helps
n = m/M = 2/18 = 1/9 ~0,1 mol
Answer:
[HI] = 0.7126 M
Explanation:
Step 1: Data given
Kc = 54.3
Temperature = 703 K
Initial concentration of H2 and I2 = 0.453 M
Step 2: the balanced equation
H2 + I2 ⇆ 2HI
Step 3: The initial concentration
[H2] = 0.453 M
[I2] = 0.453 M
[HI] = 0 M
Step 4: The concentration at equilibrium
[H2] = 0.453 - X
[I2] = 0.453 - X
[HI] = 2X
Step 5: Calculate Kc
Kc = [Hi]² / [H2][I2]
54.3 = 4x² / (0.453 - X(0.453-X)
X = 0.3563
[H2] = 0.453 - 0.3563 = 0.0967 M
[I2] = 0.453 - 0.3563 = 0.0967 M
[HI] = 2X = 2*0.3563 = 0.7126 M
