# Question b019f

Aug 16, 2016

${\text{0.18 L Na"_3"PO}}_{4}$

#### Explanation:

The balanced chemical equation that describes this double replacement reaction looks like this

$\textcolor{b l u e}{3} {\text{CaCl"_ (2(aq)) + color(purple)(2)"Na"_ 3"PO"_ (4(aq)) -> "Ca"_ 3("PO"_ 4)_ (2(s)) darr + 6"NaCl}}_{\left(a q\right)}$

In terms of moles of reactants, the balanced chemical equation tells you that for every $\textcolor{b l u e}{3}$ moles of calcium chloride present in solution, the reaction will consume $\textcolor{p u r p \le}{2}$ moles of sodium phosphate.

The problem provides you with the molarity and volume of the calcium chloride solution, which is equivalent so saying that you have the number of moles of calcium chloride present in solution

0.500 color(red)(cancel(color(black)("L solution"))) * overbrace("0.400 moles CaCl"_2/(1color(red)(cancel(color(black)("L solution")))))^(color(darkgreen)("= 0.400 M CaCl"_2)) = "0.200 moles CaCl"_2

This means that in order for all the moles of calcium chloride to react, you need to provide

0.200 color(red)(cancel(color(black)("moles CaCl"_2))) * (color(purple)(2)color(white)(a)"moles Na"_3"PO"_4)/(color(blue)(3)color(red)(cancel(color(black)("moles CaCl"_2)))) = 0.400/3color(white)(a)"moles Na"_3"PO"_4

Finally, use the molarity of the sodium phosphate solution to calculate the volume of solution that would contain this many moles of sodium phosphate

0.400/3 color(red)(cancel(color(black)("moles Na"_3"PO"_4))) * overbrace("1 L solution"/(0.75 color(red)(cancel(color(black)("moles Na"_3"PO"_4)))))^(color(darkgreen)("= 0.75 M Na"_3"PO"_4)) = "0.17778 L"#

Rounded to two sig figs, the number of sig figs you have for the molarity of the sodium phosphate solution, the answer will be

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{volume of Na"_3"PO"_4 = "0.18 L}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$