Question

30.0 mL of Na3PO4(aq) is reacted with excess 2.00 M Ba(NO3)2(aq). A precipitate of Ba3(PO4)2(s) is formed. The precipitate is filtered and dried to constant mass. The mass of precipitate produced is 4.10 g. (a) Write the balanced net ionic equation of the reaction. (b) How many moles of Na3PO4 reacted

Answer 1

Answer:

(a) $(PO_4)^{3-}(aq)+3Ba^{2+}(aq)\rightarrow Ba_3(PO_4)_2(s)$

(b) 0.0136 moles.

Explanation:

Hello!

(a) In this case, for the described chemical reaction and the requirement of the net ionic equation, we first need the complete molecular one:

$2Na_3(PO_4)(aq)+3Ba(NO_3)_2(aq)\rightarrow Ba_3(PO_4)_2(s)+6NaNO_3(aq)$

Now we split up the aqueous species into ions:

$6Na^++(PO_4)^{3-}+3Ba^{2+}+6(NO_3)^-\rightarrow Ba_3(PO_4)_2(s)+6Na^++6(NO_3)^-$

It means we can cancel out both sodium and nitrate ions are they are the spectator ones, in order to get the net ionic equation:

$(PO_4)^{3-}(aq)+3Ba^{2+}(aq)\rightarrow Ba_3(PO_4)_2(s)$

(b) Now, since 4.10 grams of barium phosphate precipitate is produced, we can compute the moles of sodium phosphate that reacted by using the molar mass of the barium phosphate and the 1:2 mole ratio between them:

$n_{Na_3PO_4}=4.10gBa_3(PO_4)_2*\frac{1molBa_3(PO_4)_2}{601.93gBa_3(PO_4)_2} *\frac{2molNa_3PO_4}{1molBa_3(PO_4)_2} \\\\n_{Na_3PO_4}=0.0136molNa_3PO_4$

Moreover, the molarity of such solution was:

$M_{Na_3PO_4}=\frac{0.0136mol}{0.0300L} =0.454M$

Best regards!