Chemistry 9th Edition

Published by Cengage Learning
ISBN 10: 1133611095
ISBN 13: 978-1-13361-109-7

Chapter 4 - Types of Chemical Reactions and Solution Stoichiometry - Exercises - Page 183: 56


There are necessary 0.250 L of that $Pb(NO_3)_2$ solution to precipitate all the lead(II) ions from that other solution.

Work Step by Step

- Write the reaction between $Na_3PO_4$ and $Pb(NO_3)_2$: $Na_3PO_4(aq) + Pb(NO_3)_2(aq) -- \gt NaNO_3 + Pb_3(PO_4)_2$ - Balance the reaction. Begin with the number of lead (Pb) atoms: $Na_3PO_4(aq) + 3Pb(NO_3)_2(aq) -- \gt NaNO_3 + Pb_3(PO_4)_2$ - Balance the number of $P$ atoms: $2Na_3PO_4(aq) + 3Pb(NO_3)_2(aq) -- \gt NaNO_3 + Pb_3(PO_4)_2$ - Balance the $Na$: $2Na_3PO_4(aq) + 3Pb(NO_3)_2(aq) -- \gt 6NaNO_3 + Pb_3(PO_4)_2$ - The equation is balanced. 1. According to the balanced equation, 2 moles of $Na_3PO_4$ reacts with 3 moles of $Pb(NO_3)_2$. 2. 1000 mL = 1 L --------------------------- Now, use these informations and the data on the exercise as conversion factors to find the $Na_3PO_4$ required volume. $150.0mL \times \frac{1L}{1000mL} \times \frac{0.250mol(Pb(NO_3)_2)}{1L} \times \frac{2mol(Na_3PO_4)}{3mol(Pb(NO_3)_2)} \times \frac{1L (Solution(Na_3PO_4))}{0.100mol(Na_3PO_4)} = 0.250L(Solution(Na_3PO_4))$
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