Strontium chloride reacts with lithium phosphate to produce strontium phosphate and lithium chloride. What mass (in grams) of lithium chloride forms when 218.3 g of lithium phosphate is used (strontium chloride is in excess). Hint: Write a balanced chemical equation.

239.8 g

Explanation:

This is a problem which involves stoichiometry calculations.

Every time we are to make a calculation based on a chemical reaction we do need to have it balanced:

SrCl₂ + Li₃PO₄ ⇒ LiCl + Sr₃ (PO₄) ₂ (Double decomposition reaction)

We need two phophate in the reactants and that will give us 6 Li we balance in the products and the rest of the coefficients we will follow:

3 SrCl₂ + 2 Li₃PO₄ ⇒ 6 LiCl + Sr₃ (PO₄) ₂

Now we can compute the moles of Li₃PO₄ that are going to react with the excess SrCl₂.

From there we will calculate the moles of LiCl formed, and hence its mass by multiplying by its molar mass.

Molar Mass Li₃PO₄ = 115.79 g/mol

moles Li₃PO₄ = 218.3 g / 115.79 g/mol = 1.885 mol

moles LiCl formed from the stoichiometry of the reaction:

1.885 mol Li₃PO₄ x (6 mol LiCl / 2 mol Li₃PO₄) = 5.656 mol Li₃PO₄

molar mass LiCl = 42.39 g/mol

mass LiCl formed:

5.656 mol x 42.39 g/mol = 239.8 g

(rounded to 4 significant figures since we had 4 significant figures in the mass of 218.3 g of Li₃PO₄)