Question #79a24

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Question #79a24

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Answer 1 · 2015-10-25T19:28:36

${1 g CaCl}_{2} \div {0.985 g Na}_{3} {PO}_{4}$ $"1 g CaCl"_2 div "0.985 g Na"_3"PO"_4$

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, ${CaCl}_{2}$ $"CaCl"_2$ , will react with trisodium phosphate, ${Na}_{3} {PO}_{4}$ $"Na"_3"PO"_4$ , to produce calcium phosphate , ${Ca}_{3} {({PO}_{4})}_{2}$ $"Ca"_3("PO"_4)_2$ , whic will precipitate out of solution, and sodium chloride, $NaCl$ $"NaCl"$ , accoding to the balanced chemical equation

$3 {CaCl}_{2(aq]} + 2 {Na}_{3} {PO}_{4(aq]} \to {Ca}_{3} {({PO}_{4})}_{2(s]} ⏐ ⏐ ↓ + 6 {NaCl}_{(aq]}$ $color(red)(3)"CaCl"_text(2(aq]) + color(green)(2)"Na"_3"PO"_text(4(aq]) -> "Ca"_3("PO"_4)_text(2(s]) darr + 6"NaCl"_text((aq])$

Notice that you have a $3 : 2$ $color(red)(3):color(green)(2)$ 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 $x$ $x$ moles of calcium chloride, you will need $\frac{2}{3} x$ $2/3x$ moles of trisodium phosphate in order to make sure that every mole of calcium chloride reacts.

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

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

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Question #79a24

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