# How do we represent stoichiometric reaction between strontium chloride, and sodium phosphate?

##### 1 Answer
Sep 9, 2017

You use the molar ratio from the balanced equation.

#### Explanation:

Your balanced equation might look something like this:

$\text{3SrCl"_2·6"H"_2"O" + "2Na"_3"PO"_4 → "Sr"_3("PO"_4)_2 + "6NaCl" + 6"H"_2"O}$

The important thing is that atoms of $\text{Sr}$ must be balanced.

The molar ratio is then either $\left(\text{1 mol Sr"_3("PO"_4)_2)/("3 mol SrCl"_2·6"H"_2"O}\right)$ or ("3 mol SrCl"_2·6"H"_2"O")/("1 mol Sr"_3("PO"_4)_2)

For example, to convert 1.20 mol of $\text{SrCl"_2·6"H"_2"O}$ to moles of $\text{Sr"_3("PO"_4)_2 }$, you would write

"1.20 mol SrCl"_2·6"H"_2"O" × ("1 mol Sr"_3("PO"_4)_2)/("3 mol SrCl"_2·6"H"_2"O") = "0.400 mol Sr"_3("PO"_4)_2