Question

__MgCl2 +

Answer 1

Answer:

24.5 g of NaCl

Explanation:

We begin from the balanced reaction:

3MgCl₂ +   2Na₃PO₄  →    6NaCl   +  Mg₃(PO₄)₂

If the sodium phosphate is in excess, then the limting reagent is the magnessium chloride.

We convert mass to moles:

20 g . 1mol / 95.2g = 0.210 moles.

3 moles of MgCl₂ can produce 6 moles of NaCl

0.210 moles of salt, may produce (0.210 . 6) /3 = 0.420 moles

Ratio of reactant is twice the product

We convert the moles to mass:

0.420 mol .  58.45 g/mol = 24.5 g