Taking into account the reaction stoichiometry, 0.18954 grams of Al(OH)₃ are formed from the reaction of 48.6 mL (0.0486 L) of 0.15 M KOH with excess Al₂(SO₄)₃.
In first place, the balanced reaction is:
Al₂(SO₄)₃ + 6 KOH → 2 Al(OH)₃ + 3 K₂SO₄
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
The molar mass of the compounds is:
Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
Molarity is a measure of the concentration of a solute in a solution and indicates the number of moles of solute that are dissolved in a given volume:
Molarity= number of moles÷ volume
In firts place, you know that 48.6 mL (0.0486 L) of 0.15 M KOH react with excess Al₂(SO₄)₃.
Replacing in the definition, you can calculate the amount of moles of KOH that react:
0.15 M= number of moles÷ 0.0486 L
Solving:
0.15 M× 0.0486 L= number of moles
0.00729 moles= number of moles
Then, 0.00729 moles of KOH react with excess Al₂(SO₄)₃. So, following rule of three can be applied: if by reaction stoichiometry 6 moles of KOH form 156 grams of Al(OH)₃, 0.00729 moles of KOH form how much mass of Na₂SO₄?
[tex]mass of Al(OH)_{3} =\frac{0.00729 moles of KOHx156 grams of Al(OH)_{3}}{6 moles of KOH}[/tex]
mass of Al(OH)₃= 0.18954 grams
Then, 0.18954 grams of Al(OH)₃ are formed from the reaction of 48.6 mL (0.0486 L) of 0.15 M KOH with excess Al₂(SO₄)₃.
Learn more about
the reaction stoichiometry:
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molarity:
brainly.com/question/9324116
brainly.com/question/10608366
brainly.com/question/7429224
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