An amount of 0.3 mole of SrCl₂ is mixed with 0.2 mole of K₃PO₄. The maximum moles of KCl which may form is;

First write the double‑replacement reaction that occurs when aqueous solutions of strontium chloride and potassium phosphate are mixed.

$3 SrCl$₂ + 2 K₃PO₄ → Sr₃(PO₄)₂ + 6 KCl

From the balanced equation we see the stoichiometric relationship:

• 3 mol SrCl₂ produce 6 mol KCl → 2 mol KCl are formed per 1 mol SrCl₂.
• 2 mol K₃PO₄ are required for 3 mol SrCl₂, i.e., the ratio SrCl₂ : K₃PO₄ = 3 : 2.

Calculate how many moles of each reactant are needed to completely react with the other.

For the given 0.3 mol of SrCl₂, the amount of K₃PO₄ required is:

$n\_\{\backslash text\{K₃PO₄,\; required\}\}\; =\; \backslash frac\{2\}\{3\}\backslash times\; 0.3\; =\; 0.2\backslash \; \backslash text\{mol\}$

The problem provides exactly 0.2 mol of K₃PO₄, so the two reactants are present in the exact stoichiometric proportion. Neither is in excess; both will be completely consumed.

Now determine the amount of KCl produced from the 0.3 mol of SrCl₂.

$n\_\{\backslash text\{KCl\}\}\; =\; 2\; \backslash times\; n\_\{\backslash text\{SrCl₂\}\}\; =\; 2\; \backslash times\; 0.3\; =\; 0.6\backslash \; \backslash text\{mol\}$

Therefore, the maximum amount of potassium chloride that can form is 0.6 mol.