Question

Calculate the mass of srontium phosphate that forms when 6.25 grams of strontium chloride react with excess phosphoric acid (H₃PO₄). Hydrochloric acid (HCI) is the other produce of this reaction. Be sure to write and balance the reaction first.

Answer 1

Molar mass of SrCl2,

MM = 1*MM(Sr) + 2*MM(Cl)

= 1*87.62 + 2*35.45

= 158.52 g/mol

mass(SrCl2)= 6.25 g

number of mol of SrCl2,

n = mass of SrCl2/molar mass of SrCl2

=(6.25 g)/(158.52 g/mol)

= 3.943*10^-2 mol

Balanced chemical equation is:

3 SrCl2 + 2 H3PO4 ---> Sr3(PO4)2 + 6 HCl

Molar mass of Sr3(PO4)2,

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

= 3*87.62 + 2*30.97 + 8*16.0

= 452.8 g/mol

According to balanced equation

mol of Sr3(PO4)2 formed = (1/3)* moles of SrCl2

= (1/3)*0.0394

= 0.0131 mol

mass of Sr3(PO4)2 = number of mol * molar mass

= 1.314*10^-2*4.528*10^2

= 5.95 g

Answer: 5.95 g