Question

Given the balanced equation:

2Na3PO4 + 3MgCl2 —> 6NaCl + Mg3(PO4)2

How many grams of Mg3(PO4)2 will be produced from 100.0mL of 0.200M Na3PO4 and excess MgCl2?

How many formula units of MgCl2 will react with 4.4 moles of Na3PO4?

Answer 1

1)

Balanced chemical equation is:

2Na3PO4 + 3MgCl2 —> 6NaCl + Mg3(PO4)2

lets calculate the mol of Na3PO4

volume , V = 100 mL

= 0.1 L

use:

number of mol,

n = Molarity * Volume

= 0.2*0.1

= 2*10^-2 mol

According to balanced equation

mol of Mg3(PO4)2 FORMED = (1/2)* moles of Na3PO4

= (1/2)*2*10^-2

= 1*10^-2 mol

This is number of moles of Mg3(PO4)2

Molar mass of Mg3(PO4)2,

MM = 3*MM(Mg) + 2*MM(P) + 8*MM(O)

= 3*24.31 + 2*30.97 + 8*16.0

= 262.87 g/mol

use:

mass of Mg3(PO4)2,

m = number of mol * molar mass

= 1*10^-2 mol * 2.629*10^2 g/mol

= 2.629 g

Answer: 2.63 g

2)

moles of MgCl2 reacted = (3/2)*number of moles of Na3PO4

= (3/2)*4.4 mol

= 6.6 mol

Number of MgCl2 = number of mol * Avogadro’s number

= 6.6 * 6.022*10^23

= 4.0*10^24 units

Answer: 4.0*10^24 units