A student mixes 125.0 mL of 0.0250 M phosphoric acid (aq) with 75.0 mL of 0.0500

Question

A student mixes 125.0 mL of 0.0250 M phosphoric acid (aq) with 75.0 mL of 0.0500 M calcium hydroxide (aq), and an observable eqaction occurs.
i) write the balanced molecular equation for the reaction.

ii) write the balanced net ionic equation for the reaction.

iii) identify the spectator ions (if any)

iv) identify the limiting reactant

v) calculate the theoretical yield (mass) of precipitate.

vi) Identify the liquid product and calculate the theoretical yield of liquid product (mass and volume)

vii) If te student's percent yield of product is only 45.0%, calculate his actual yield of precipitate.

viii) calculate the volume of excess reactant that remains unreacted.

Explanation / Answer

1) 2H3PO4 + 3Ca(OH)2 --> Ca3(PO4)2(s) + 6HOH(l)

2)(3Ca2+)aq + (6OH-)aq + (6H1+)aq + (2PO4,3-)aq ---> (6H2O)l + (3Ca2+) +2PO4,3-)s

3) spectator ions: OH^- and H^+

4) Number of moles of Ca(OH)2 ->0.0500*1000/75 = 0.6667

Number of moles of H3PO4 ->0.0250*1000/125 = 0.2 moles

Limiting reagent is H3PO4

5) Yield in mass(Percentage) = 0.2 moles of H3PO4 will produce 0.1 mole of Ca3(PO4)2

hence mass produced = 0.1 * 310 = 31 gms of salt is produced

6)Yield in mass(Percentage) = 0.2 moles of H3PO4 will produce 0.6 mole of H20

hence mass produced = 0.6 * 18 = 10.8 gms of liquid is produced

7)Actual yield of Precipitate = 45/100*31 = 13.95 gm

8) Volume of excess reagent that left unreacted = 0.36667 moles of Ca(OH)2

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