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For the following reaction, 76.0 grams of barium chloride are allowed to react with 67.0 grams of potassium sulfate.
The reaction consumes _____ moles of barium chloride. The reaction produces _____ moles of barium sulfate and _____ moles of potassium chloride.
Answer: a) The reaction consumes 0.365 moles of barium chloride.
b) The reaction produces 0.365 moles of barium sulfate and 0.730 moles of potassium chloride.
Explanation:
To calculate the moles :
According to stoichiometry :
1 mole of require 1 mole of
Thus 0.365 moles of will require=
of
Thus is the limiting reagent as it limits the formation of product and
is the excess reagent.
As 1 moles of give = 1 moles of
Thus 0.365 moles of give =
of
As 1 moles of give = 2 moles of
Thus 0.365 moles of give =
of
Thus the reaction consumes 0.365 moles of barium chloride. The reaction produces 0.365 moles of barium sulfate and 0.730 moles of potassium chloride.