Respuesta :
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!!!
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 [tex]\frac{5.943 \times 10^{24}}{6.02 \times 10^{23}} moles = 9.869 \, moles \, of \, H_3PO_4[/tex]
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.
