Starting with the following equation, MgCl₂(aq) + Na₃PO₄(aq) → Mg₃(PO₄)₂(s) + NaCl(aq) calculate the moles of MgCl₂ that will be required to produce 545 grams of Mg₃(PO₄)₂.

Answer 1

Start with a correctly balanced equation:

3MgCl₂(aq) + 2Na₃PO₄(aq) → Mg₃(PO₄)₂(s) + 6NaCl(aq) ... balanced equation

Use dimensional analysis and stoichiometry of the balanced equation to find moles of MgCl₂ needed:

molar mass Mg₃(PO₄)₂ = 263 g / mol

545 g Mg₃(PO₄)₂ x 1 mol / 263 g x 3 mol MgCl₂ / 1 mol Mg₃(PO₄)₂ = 6.22 mols MgCl₂ needed

Answer 2

Mole of Mgz (PO4)2 = given mous → 535g → 2 mol. molar mais 262.8glmd

1ot write the Balanced equation :

3 MgCl2 (09) +2 Naz PO4 (aq) ->Mg 3 (PO215 + 6 Nack (0) (5)

☆ From Balanced equation lle Can Conclude:

Imole At Mg₂ (POL)z produce when - 3 mole Mycy jeact... So, 2 mole Mg 3 (PO4)₂ produce when • (2X3) mole Mych roact

2 mole Mgz (PO4)₂ produce when - 6 mole MgCl₂ react.

Mole op MgCl₂ = 6 mdl