Taking into account the reaction stoichiometry and molarity, 10.17 mL of 0.32 M H₂C₂O₄ is required to titrate 15.50 mL of 0.21 M Mg(OH)₂.
In first place, the balanced reaction is:
H₂C₂O₄ + Mg(OH)₂ → MgC₂O₄ + 2 H₂O
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:
Molar concentration or 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.
The molarity of a solution is calculated by dividing the moles of solute by the volume of the solution:
[tex]Molarity=\frac{number of moles}{volume}[/tex]
Molarity is expressed in units [tex]\frac{moles}{liter}[/tex].
In this case, you know:
Replacing in the definition of molarity:
[tex]0.21 M=\frac{number of moles}{0.0155 L}[/tex]
Solving:
number of moles= 0.21 M× 0.0155 L
number of moles= 0.003255 moles
So, you must titrate 0.003255 moles of Mg(OH)₂.
The following rule of three can be applied: If by reaction stoichiometry 1 mole of Mg(OH)₂ react with 1 mole of H₂C₂O₄, 0.003255 moles of Mg(OH)₂ react with how many moles of H₂C₂O₄?
[tex]amount of moles of H_{2} C_{2} O_{4} =\frac{0.003255 moles of Mg(OH)_{2} x1 mole of H_{2} C_{2} O_{4}}{1mole of Mg(OH)_{2}}[/tex]
amount of moles of H₂C₂O₄= 0.003255 moles
0.003255 moles of H₂C₂O₄ is required to titrate 15.50 mL of 0.21 M Mg(OH)₂.
In this case, you know:
Replacing in the definition of molarity:
[tex]0.32 M=\frac{0.003255 M}{volume}[/tex]
Solving:
0.32 M× volume= 0.003255 moles
volume= 0.003255 moles÷ 0.32 M
volume= 0.01017 L= 10.17 mL (being 1 L= 1000 mL)
Finally, 10.17 mL of 0.32 M H₂C₂O₄ is required to titrate 15.50 mL of 0.21 M Mg(OH)₂.
Learn more about
the reaction stoichiometry:
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brainly.com/question/24653699
molarity:
brainly.com/question/9324116
brainly.com/question/10608366
brainly.com/question/7429224
