Question

Обчисліть масу й кількість ортофосфатної кислоті, необхідної для цілковитої нейтралізації трикислотної основи кількістю речовини 0.5 моль.

Answer 1

Решение есть на фотографии

Answer 2

Calculation of the Mass and Amount of Orthophosphoric Acid

To calculate the mass and amount of orthophosphoric acid required for the complete neutralization of a triacidic base with a quantity of 0.5 moles, we need to use the balanced chemical equation for the reaction between the acid and the base.

The balanced chemical equation for the reaction between orthophosphoric acid (H₃PO₄) and a triacidic base can be represented as follows:

H₃PO₄ + 3M(OH)₃ → M₃PO₄ + 3H₂O

In this equation, M represents the cation of the triacidic base.

To determine the mass and amount of orthophosphoric acid required, we need to consider the stoichiometry of the reaction. From the balanced equation, we can see that one mole of orthophosphoric acid reacts with three moles of the triacidic base.

Given that the quantity of the triacidic base is 0.5 moles, we can calculate the amount of orthophosphoric acid required as follows:

Amount of orthophosphoric acid = 0.5 moles × (1 mole H₃PO₄ / 3 moles triacidic base)

Using this calculation, we find that the amount of orthophosphoric acid required is approximately 0.1667 moles.

To calculate the mass of orthophosphoric acid required, we need to know the molar mass of H₃PO₄. The molar mass of H₃PO₄ is approximately 98 g/mol.

Mass of orthophosphoric acid = Amount of orthophosphoric acid × Molar mass of H₃PO₄

Substituting the values, we get:

Mass of orthophosphoric acid = 0.1667 moles × 98 g/mol

Calculating this, we find that the mass of orthophosphoric acid required is approximately 16.33 grams.

Therefore, to completely neutralize the triacidic base with a quantity of 0.5 moles, approximately 0.1667 moles (or 16.33 grams) of orthophosphoric acid is needed.

Please note that the sources provided did not contain specific information related to the calculation. The calculation was performed based on the balanced chemical equation and general knowledge of stoichiometry.