Final answer:
When 455 g of Pb(NO3)2 and 215 g of K3PO4 are mixed, 370.93 g of Pb3(PO4)2 precipitate is expected to form based on stoichiometry and the given balanced reaction.
Step-by-step explanation:
The student asked what mass, in grams, of Pb₃(PO₄)₂ precipitate will be formed when 455 g of Pb(NO₃)₂ and 215 g of K₃PO₄ are mixed according to the reaction:
3 Pb(NO₃)₂(aq) + 2 K₃PO₄(aq) → Pb₃(PO₄)₂(s) + 6 KNO₃(aq)
To solve this, first we must calculate the number of moles of both reactants:
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- Pb(NO₃)₂: 455 g / 331.21 g/mol = 1.37 moles
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- K₃PO₄: 215 g / 212.27 g/mol = 1.01 moles
Using the reaction stoichiometry, we can see that 3 moles of Pb(NO₃)₂ react with 2 moles of K₃PO₄ to produce 1 mole of Pb₃(PO₄)₂. Pb(NO₃)₂ is the limiting reactant, so we use its moles to determine the amount of Pb₃(PO₄)₂:
1.37 moles Pb(NO₃)₂ (1 mole Pb₃(PO₄)₂ / 3 moles Pb(NO₃)₂) = 0.457 moles Pb₃(PO₄)₂
0.457 moles Pb₃(PO₄)₂ × 811.54 g/mol = 370.93 g of Pb₃(PO₄)₂
Therefore, 370.93 g of Pb₃(PO₄)₂ precipitate can be expected to form.