Question

Question #79a24

Answer 1

`"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, `"CaCl"_2`, will react with trisodium phosphate, `"Na"_3"PO"_4`, to produce calcium phosphate ,`"Ca"_3("PO"_4)_2`, whic will precipitate out of solution, and sodium chloride, `"NaCl"`, accoding to the balanced chemical equation

`color(red)(3)"CaCl"_text(2(aq]) + color(green)(2)"Na"_3"PO"_text(4(aq]) -> "Ca"_3("PO"_4)_text(2(s]) darr + 6"NaCl"_text((aq])`

Notice that you have a `color(red)(3):color(green)(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 `2/3x` 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 `3/2x` 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 `"3 moles"` of calcium chloride and `"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

`"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 `"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

`"1 g CaCl"_2 div "0.985 g Na"_3"PO"_4`