0.340 g or BaCl 2 x 2H 2 O and0.624g of Na 3 PO 4 x 12H 2 O are dissolved in 500

Question

0.340 g or BaCl2 x 2H2O and0.624g of Na3PO4 x 12H2Oare dissolved in 500mL of water to form asolution.
a. Which is the limiting reactant?
b. How many grams of Ba3(PO4)2 willprecipitate? c. How many grams of the excess reactantremain after the reaction has gone tocompletion?
*WILL RATE LIFESAVER IF ANSWER EACH PART* 0.340 g or BaCl2 x 2H2O and0.624g of Na3PO4 x 12H2Oare dissolved in 500mL of water to form asolution.
b. How many grams of Ba3(PO4)2 willprecipitate? c. How many grams of the excess reactantremain after the reaction has gone tocompletion?
*WILL RATE LIFESAVER IF ANSWER EACH PART*

Explanation / Answer

n(BaCl2 x 2H2O) = m/M = 0.340/244.232 =0.001392mol n(Na3PO4 x 12H2O) = m/M =0.624/380.132 = 0.001642mol 3BaCl2 + 2Na3PO4 -> 6NaCl +Ba3(PO4)2 Stoichiometry ratio =n(BaCl2)/n(Na3PO4) = 3/2 = 1.5 Actual ratio = 0.001392/0.001642 = 0.8477 Stoichiometry ratio > Actual ratio So, BaCl2 is the limiting reagent n( Ba3(PO4)2) = 1/3 xn(BaCl2) = 1/3 x 0.001392 = 0.000464mol m(Ba3(PO4)2) = nM = 0.000464 x601.84 = 0.279g n(excess reactant) = n(excess Na3PO4) =0.001642 - 2/3 x 0.001392 = 0.000714mol m(Na3PO4) = nM = 0.000714 x 163.94 =0.117g Hope this helps!

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