Determine the molality and molarity of a sodium chloride solution prepared by adding 52.80 g of solid NaCl to 150.0 g of water.
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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