g aqueous barium hydroxide (ba(oh)2) and nitric acid (hno3) participate in a complete neutralization reaction. in the molecular
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<u>Answer:</u> The mass of potassium sulfate that can be produced is 73.88 grams
<u>Explanation:</u>
To calculate the number of moles, we use the equation:
.....(1)
- <u>For KOH:</u>
Given mass of KOH = 23.8 g
Molar mass of KOH = 56.1 g/mol
Putting values in equation 1, we get:
The chemical equation for the reaction of KOH and potassium hydrogen sulfate follows:
As, potassium hydrogen sulfate is present in excess. It is considered as an excess reagent.
KOH is considered as a limiting reagent because it limits the formation of product.
By Stoichiometry of the reaction:
1 mole of KOH produces 1 mole of potassium sulfate
So, 0.424 moles of KOH will produce = of potassium sulfate
Now, calculating the mass of potassium sulfate from equation 1, we get:
Molar mass of potassium sulfate = 174.26 g/mol
Moles of potassium sulfate = 0.424 moles
Putting values in equation 1, we get:
Hence, the mass of potassium sulfate that can be produced is 73.88 grams