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Assume 0.18 L of a 5.0 M solution of lead (II) nitrate, Pb(NO3)2, reacts with a 2.6 M solution of sodium phosphate, Na3PO4, to produce lead (II) phosphate, Pb3(PO4)2, and sodium nitrate, NaNO3. The problem requires that you determine the volume of sodium phosphate, Na3PO4, needed for the reaction to occur.

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dsdrajlin

Answer:

0.23 L

Explanation:

Let's consider the following balanced equation.

3 Pb(NO₃)₂(aq) + 2 Na₃PO₄(aq) ⇄ Pb₃(PO₄)₂(s) + 6 NaNO₃(aq)

The moles of Pb(NO₃)₂ are:

[tex]0.18L\times \frac{5.0mol}{L} =0.90mol[/tex]

The molar ratio of Pb(NO₃)₂ to Na₃PO₄ is 3:2. The moles of Na₃PO₄ are:

[tex]0.90molPb(NO_{3})_{2}.\frac{2molNa_{3}PO_{4}}{3molPb(NO_{3})_{2}} =0.60molNa_{3}PO_{4}[/tex]

The volume of Na₃PO₄ required is:

[tex]\frac{0.60mol}{2.6mol/L} =0.23L[/tex]

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