Question 79a24

Oct 25, 2015

${\text{1 g CaCl"_2 div "0.985 g Na"_3"PO}}_{4}$

Explanation:

Since you didn't provide the actual data you obtained, I will show you what the ratio should come out to be.

Calcium chloride, ${\text{CaCl}}_{2}$, will react with trisodium phosphate, ${\text{Na"_3"PO}}_{4}$, to produce calcium phosphate ,"Ca"_3("PO"_4)_2, whic will precipitate out of solution, and sodium chloride, $\text{NaCl}$, accoding to the balanced chemical equation

$\textcolor{red}{3} {\text{CaCl"_text(2(aq]) + color(green)(2)"Na"_3"PO"_text(4(aq]) -> "Ca"_3("PO"_4)_text(2(s]) darr + 6"NaCl}}_{\textrm{\left(a q\right]}}$

Notice that you have a $\textcolor{red}{3} : \textcolor{g r e e n}{2}$ mole ratio between calcium chloride and trisodium phosphate. This means that the reaction wil lalways consume the two reactants in this mole ratio.

If you start with $x$ moles of calcium chloride, you will need $\frac{2}{3} x$ moles of trisodium phosphate in order to make sure that every mole of calcium chloride reacts.

LIkewise, if you start with $y$ moles of trisodium phosphate, you will need $\frac{3}{2} x$ moles of calcium chloride.

Now, since you gave no information about the masses of the two reactants you started with, I'll assume that you had $\text{3 moles}$ of calcium chloride and $\text{2 moles}$ of trisodium phosphate.

Use the molar masses of the two compounds to figure out what masses of each you'd have

3color(red)(cancel(color(black)("moles CaCl"_2))) * "110.984 g"/(1color(red)(cancel(color(black)("mole CaCl"_2)))) = "332.95 g CaCl"_2

and

2color(red)(cancel(color(black)("moles Na"_3"PO"_4))) * "163.941 g"/(1color(red)(cancel(color(black)("mole Na"_3"PO"_4)))) = "327.88 g Na"_3"PO"_4

This means that you can convert the $3 : 2$ mole ratio to gram ratio by dividing both masses by the smallest one to get

$\text{For CaCl"_2: " "(332.95color(red)(cancel(color(black)("g"))))/(327.88color(red)(cancel(color(black)("g")))) = "1.0155}$

"For Na"_3"PO"_4: " "(327.88color(red)(cancel(color(black)("g"))))/(327.88color(red)(cancel(color(black)("g")))) = 1

SInce you need to express this as $\text{1 g}$ of calcium chloride to grams of trisodium phosphate, you can say that

1color(red)(cancel(color(black)("g CaCl"_2))) * ("1 g Na"_3"PO"_4)/(1.0155color(red)(cancel(color(black)("g CaCl"_2)))) = "0.985 g Na"_3"PO"_4#

Therefore, th gram ratio will be

${\text{1 g CaCl"_2 div "0.985 g Na"_3"PO}}_{4}$