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A 17.50-ml sample of a 0.150 m solution of h3po4 reacts with excess ba(oh)2. what mass of h2o is produced in the reaction?

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CintiaSalazar

Taking into account the reaction stoichiometry, 0.14175 grams of H₂O are formed when a 17.50 mL sample of a 0.150 M solution of H₃PO₄ reacts with excess Ba(OH)₂.

Reaction stoichiometry

In first place, the balanced reaction is:

2 H₃PO₄ + 3 Ba(OH)₂  → Ba₃(PO₄)₂ + 6 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:

  • H₃PO₄: 2 moles
  • Ba(OH)₂: 3 moles
  • Ba₃(PO₄)₂: 1 mole
  • H₂O: 6 moles

The molar mass of the compounds is:

  • H₃PO₄: 98 g/mole
  • Ba(OH)₂: 171.34 g/mole
  • Ba₃(PO₄)₂: 602 g/mole
  • H₂O: 18 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

  • H₃PO₄: 2 moles ×98 g/mole= 196 grams
  • Ba(OH)₂: 3 moles ×171.34 g/mole= 514.02 grams
  • Ba₃(PO₄)₂: 1 mole ×602 g/mole= 602 grams
  • H₂O: 6 moles ×18 g/mole= 108 grams

Moles of H₃PO₄ required

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:

molarity= amount of moles÷ volume

In this case, you know for the reactant H₃PO₄:

  • molarity= 0.150 M
  • volume= 17.50 mL= 0.0175 L (being 1000 mL= 1 L)

Replacing in the definition of molarity:

0.150 M= amount of moles÷ 0.0175 L

0.150 M× 0.0175 L= amount of moles

0.002625 moles= amount of moles

So, 0.002625 moles of H₃PO₄ react.

Mass of H₂O formed

The following rules of three can be applied: if by reaction stoichiometry 2 moles of H₃PO₄ form 108 grams of H₂O, 0.002625 H₃PO₄ form how much mass of H₂O?

[tex]mass of H_{2}O=\frac{0.002625 moles of H_{3}PO_{4} x108 grams of H_{2}O}{2moles of H_{3}PO_{4}}[/tex]

mass of H₂O= 0.14175 grams

Finally, 0.14175 grams of H₂O are formed when a 17.50 mL sample of a 0.150 M solution of H₃PO₄ reacts with excess Ba(OH)₂.

Learn more about the reaction stoichiometry:

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molarity:

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