Question

In: Chemistry

You are given 50.00 mL of 0.699 ammonium phosphate and 50.00 mL of 0.888 M lead...

You are given 50.00 mL of 0.699 ammonium phosphate and 50.00 mL of 0.888 M lead nitrate. Will a precipitate form? If so, write the balanced molecular reaction equation and the balanced net ionic reaction equation, then calculate the theoretical yield (in grams) of the product formed

Expert Solution

The reaction of ammonium phosphate with lead nitrate is

Molecular equation

2(NH₄)₃PO₄ + 3Pb(NO₃)₂ ---> 6NH₄NO₃ + Pb₃(PO₄)₂ (s)

Net ionic reaction

2PO₄^-3 + 3Pb⁺²---> Pb₃(PO₄)₂ (s)

There will be formation of ppt of lead phosphate

Moles of ammonium phosphate taken = Molarity X volume = 0.699 X 50 millimoles = 34.95millimoles

Moles of lead nitrate = Molarity X volume = 0.888 X 50 millimoles = 44.4 millimoles

As per balanced equation two moles of ammonium phosphate will react with three moles of lead nitrate to give one mole of Pb₃(PO₄)₂

Here 34.95 millimoles of ammonium phosphate will required = 1.5 X 34.95 millimoles of lead nitrate

= 52.43 millimoles

so limiting reagent is lead nitrate

Now, three moles of lead nitrate will give one mole of lead phosphate precipitate

therefore 44.4 millimoles of lead nitrate will give = 44.4/3 millimoles of lead phosphate = 14.8 millimoles of lead phosphate

Molecular weight of lead phosphate = 811.54 g/mol

MAss of lead phosphate (in grams) obtained = Moles X molecular weight

= 14.8 millimoles X 811.54 g/mole = 12.011 grams

queen_honey_blossom answered 2 years ago

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