55 grams 3.23gml what's the volume
1 answer:
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Answer : The mass of ammonia produced can be, 121.429 k
Solution : Given,
Mass of = 100 kg = 100000 g
Mass of = 100 kg = 100000 g
Molar mass of = 28 g/mole
Molar mass of = 2 g/mole
Molar mass of = 17 g/mole
First we have to calculate the moles of and
.
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
From the balanced reaction we conclude that
As, 1 mole of react with 3 mole of
So, 3571.43 moles of react with
moles of
From this we conclude that, is an excess reagent because the given moles are greater than the required moles and
is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of
From the reaction, we conclude that
As, 1 mole of react to give 2 mole of
So, 3571.43 moles of react to give
moles of
Now we have to calculate the mass of
Therefore, the mass of ammonia produced can be, 121.429 kg
The question is incomplete, complete question is :
The frequency factors for these two reactions are very close to each other in value. Assuming that they are the same, compute the ratio of the reaction rate constants for these two reactions at 25°C.
Activation energy of the reaction 1 , = 14.0 kJ/mol
Activation energy of the reaction 2, = 11.9 kJ/mol
Answer:
0.4284 is the ratio of the rate constants.
Explanation:
According to the Arrhenius equation,
The expression used with catalyst and without catalyst is,
where,
= rate constant reaction -1
= rate constant reaction -2
Activation energy of the reaction 1 , = 14.0 kJ/mol = 14,000 J
Activation energy of the reaction 2, = 11.9 kJ/mol = 11,900 J
R = gas constant = 8.314 J/ mol K
T = temperature =
Now put all the given values in this formula, we get
0.4284 is the ratio of the rate constants.