Question

Calculate the theoretical yield of the precipitate when mixing 235 ml of 0.7548 M Na3PO4 with 197 mL of 1.354 M Ba(No3)2

Answer 1

2 Na₃PO₄ + 3 Ba(NO₃)₂ = 6 NaNO₃ + Ba₃(PO₄)₂
Reaction type: double replacement.

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Theoretical moles of Na₃PO₄ : Ba(NO₃)₂ = 2 : 3

We know the equation can be expressed as:

Theoretical yield = moles of solutes x molar mass of products.

So we have to find the moles of solutes for the mixing.

moles Na₃PO₄ = 0.7548 mol/1000ml * 235ml = 0.1774 moles

molar mass Na₃PO₄ = 163.94 g/mol

moles Ba(NO₃)₂ = 1.354 mol/1000ml* 197ml = 0.2667 moles

molar mass Ba(NO₃)₂ = 261.37 g/mol

Experimental moles of Na₃PO_{4 :} Ba(NO₃)2 = 0.1774 moles : 0.2667 moles = 1.50 : 1 = 2 : 1.33

Since Ba(NO₃)₂ is in deficit, we will use it as the limiting reactant

6 moles NaNO₃ = 1/3 (0.2667 moles Ba(NO₃)₂)

moles NaNO₃ = 1.48E^-2

molar mass NaNO₃ = 84.9947 g/mol

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Theoretical Yield of NaNO₃= (1.48E^-2 mol)x(84.9947 g/mol) = 1.2593 grs.

The theoretical yield of the precipitate is 1.2593 grs.