Question

__MgCl2 +

_Na3PO4 -->
NaCl
_Mg3(PO4)2 +
---
In the reaction above, how many moles of sodium chloride would be produced
if you started with 20 grams of magnesium chloride and excess sodium
phosphate? The equation is not balanced.
(If anyone can answer in depth if appreciate it )

Answer 1

24.5 g of NaCl

Step-by-step 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