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

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

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.