In a 2.2 m solution with the same volume there would be 11.825 g of a salt
Answer:
4.65 L of NH₃ is required for the reaction
Explanation:
2NH₃(g) + H₂SO₄(aq) → (NH₄)₂SO₄(s)
We determine the ammonium sulfate's moles that have been formed.
8.98 g . 1mol / 132.06 g = 0.068 moles
Now, we propose this rule of three:
1 mol of ammonium sulfate can be produced by 2 moles of ammonia
Therefore, 0.068 moles of salt were produced by (0.068 . 29) / 1 = 0.136 moles of NH₃. We apply the Ideal Gases Law, to determine the volume.
Firstly we do unit's conversions:
27.6°C +273 = 300.6 K
547.9 mmHg . 1 atm / 760 mmHg = 0.721 atm
V = ( n . R . T ) / P → (0.136 mol . 0.082 L.atm/mol.K . 300.6K) / 0.721 atm
V = 4.65 L
Answer:
K=CHANGE IN CONCENTRATION/TIME TAKEN
Explanation:
Answer:
29.75 Kg of NH₃
Solution:
In order to calculate the theoretical yield, first we will identify the limiting reactant.
According to equation,
6 g (3 moles) H₂ requires = 28 g (1 mole) N₂
So,
5250 g H₂ will require = X g of N₂
Solving for X,
X = (5250 g × 28 g) ÷ 6 g
X = 25433 g of N₂
Hence, it is found that H₂ is the limiting reactant because N₂ is provided in excess (32700 g). Therefore,
As,
6 g (3 mole) H₂ produced = 34 g ( 2 moles) of NH₃
So,
5250 g H₂ will produce = X g of NH₃
Solving for X,
X = (5250 g × 34 g) ÷ 6 g
X = 29750 g of NH₃
Or,
X = 29.75 Kg of NH₃
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