Question

5.943x10^24 molecules of H3PO4 will need how many grams of Mg(OH)2 in the reaction below? 3 Mg(OH)2 + 2 H3PO4 -------> 1 Mg3(PO4)2 + 6 H2O

Answer 1

Answer:

5.943 × 10²⁴ molecules of H₃PO₄ will react with 789.29 grams of  Mg(OH)₂

Explanation:

Here we have

5.943 × 10²⁴ molecules of H₃PO₄ in a reaction with Mg(OH)₂

as follows

3Mg(OH)₂ + 2H₃PO₄ → Mg₃(PO₄)₂ + 6H₂O

Therefore 3 moles of Mg(OH)₂ react with 2 moles of H₃PO₄ to form 1 mole of Mg₃(PO₄)₂ and 6 moles of H₂O

5.943 × 10²⁴ molecules of H₃PO₄ which is equivalent to $\frac{5.943 \times 10^{24}}{6.02 \times 10^{23}} moles = 9.869 \, moles \, of \, H_3PO_4$

Will react with 3/2×9.869 moles or 14.8 moles of Mg(OH)₂

One mole of Mg(OH)₂ weighs 58.3197 g/mol

Therefore, 5.943 × 10²⁴ molecules of H₃PO₄ will react with 14.8×58.3197 g or 789.29 grams of  Mg(OH)₂.

Answer 2

Answer:

Mass of Mg(OH)₂ required for the reaction = 863.13 g

Explanation:

3Mg(OH)₂ + 2H₃PO₄ -------> Mg₃(PO₄)₂ + 6H₂O

(5.943 x 10²⁴) molecules of H₃PO₄ is available fore reaction. Mass of Mg(OH)₂ required for reaction.

According to Avogadro's theory, 1 mole of all substances contain (6.022 × 10²³) molecules.

This can allow us find the number of moles that (5.943 x 10²⁴) molecules of H₃PO₄ represents.

1 mole = (6.022 × 10²³) molecules.

x mole = (5.943 x 10²⁴) molecules

x = (5.943 x 10²⁴) ÷ (6.022 × 10²³)

x = 9.87 moles

From the stoichiometric balance of the reaction,

2 moles of H₃PO₄ reacts with 3 moles of Mg(OH)₂

9.87 moles of H₃PO₄ will react with y moles of Mg(OH)₂

y = (3×9.87)/2 = 14.80 moles

So, 14.8 moles of Mg(OH)₂ is required for this reaction. We them convert this to mass

Mass = (number of moles) × Molar mass

Molar mass of Mg(OH)₂ = 58.3197 g/mol

Mass of Mg(OH)₂ required for the reaction

= 14.8 × 58.3197 = 863.13 g

Hope this Helps!!!