Question

Consider the reaction 3CaCl₂(aq) + 2Na₃PO₄(aq) --> Ca₃(PO₄)₂(s) + 6 NaCl(aq) (molar mass of calcium phosphate is 310.18 g/mol)

What mass of Ca₃(PO₄)₂ is produced from 275 mL of 0.150 M CaCl₂? _____ g (

Answer 1

ANSWER:

The given reaction is:

3 CaCl₂ (aq) + 2 Na₃PO₄ (aq) Ca₃(PO₄)₂ (s) + 6 NaCl (aq)

Now,

concentration of CaCl₂ = 0.150 M

volume of CaCl₂ = 275 mL = 0.275 L

number of moles of CaCl₂ = concentration x volume

= 0.150 M x 0.275 L

= 4.125 x 10^-2 mol

In the given reaction,

3 moles of CaCl₂ produced 1 mole of Ca₃(PO₄)₂

1 moles of CaCl₂ produced 1/3 mole of Ca₃(PO₄)₂

And,

4.125 x 10^-2 mol of CaCl₂ will produce = (1/3) x 4.125 x 10^-2 mol of Ca₃(PO₄)₂

= 1.375 x 10^-2 mol of Ca₃(PO₄)₂

Now,

mass of Ca₃(PO4)₂ = molecular weight x number of moles

= 310.18 g/mol x 1.375 x 10^-2 mol

= 4.26 g

Hence, mass of Ca₃(PO4)₂ is produced from 275 mL of 0.150 M CaCl₂ = 4.26 gram.