Calculate the concentration in molality of calcium phosphate in a solution containing 250.0 grams of pure water when the freezin
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Answer : The theoretical yield of potassium carbonate is, 146.483 g
The percent yield of potassium carbonate is, 85.33 %
Solution : Given,
Mass of = 150 g
Mass of = 90 g
Molar mass of = 212.27 g/mole
Molar mass of = 233.99 g/mole
Molar mass of = 138.205 g/mole
First we have to calculate the moles of and
The given balanced reaction is,
From the given reaction, we conclude that
2 moles of react with 1 mole of
0.7066 moles of react with
moles of
But the moles of is, 0.3846 moles.
So, is an excess reagent and
is a limiting reagent.
Now we have to calculate the moles of .
As, 2 moles of react to give 3 moles of
So, 0.7066 moles of react to give
moles of
Now we have to calculate the mass of .
The theoretical yield of potassium carbonate = 146.483 g
The experimental yield of potassium carbonate = 125 g
Now we have to calculate the % yield of potassium carbonate.
Formula for percent yield :
Therefore, the % yield of potassium carbonate is, 85.33%