Question #79a24

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#