What volume of 0.155M Na3PO4 solution is necessary to completely react with 90.7mL of 0.102M CuCl2?

Question

aaronq · Answer

start by writing (and balancing) the equation for the reaction

anonymous · Answer

$$2Na _{3}PO _{4}+3CuCl _{2}\rightarrow6NaCl+Cu _{3}(PO _{4})_{2}$$

anonymous · Answer

So I would start off with .00925 mol of$$CuCl _{2}$$

aaronq · Answer

for problems involving stoichiometry, you have to convert whatever information they give you to moles (signified by the symbol "n")

convert 90.7mL of 0.102M $CuCl_2$ to moles

Use the stoichiometric coefficients to build a ratio, like this:

e.g. for a general reaction:

                      $\color{red}{a}A + \color{blue}{b}B$ $\rightleftharpoons$ $ \color{green}{c}C$

where upper case are the species (A,B,C), and lower case (a,b,c) are the coefficients,

                     $\dfrac{n_A}{\color{red}{a}}=\dfrac{n_B}{\color{blue}{b}}=\dfrac{n_C}{\color{green}{c}}$

once you find the moles of $Na_2PO_4$, you can convert back to volume using the concentration given.

anonymous · Answer

Does 25.13L sound right?

aaronq · Answer

no thats way too much

anonymous · Answer

I got .0061676 mol of NaPO4

anonymous · Answer

then divided the molarity given by the previous number (moles)

aaronq · Answer

and you got 25 Litres?

anonymous · Answer

yes...

anonymous · Answer

What did I do wrong?

aaronq · Answer

$\dfrac{0.0907\;L*0.102M}{3}=\dfrac{n_{Na_3PO4}}{2}$

$n_{Na_3PO_4}=0.0604666666666667$

$L_{solution}=\dfrac{n_{Na_3PO_4}}{M_{Na_3PO_4}}=\dfrac{0.0604666666666667}{0.155}$

=0.39010 L

anonymous · Answer

Oh... I completely flipped the moles and Molarity. *facepalm* Makes more sense now! Thanks!

aaronq · Answer

no problem, hope its clear !