CHEM 107 Lecture Notes - Lecture 51: United States Environmental Protection Agency, Copper(Ii) Chloride, Silver Nitrate

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11 Aug 2017
School
Department
Course
4-3-17-Stoichiometry
Monday, April 3, 2017
10:58 AM
If 35.0 mL of 0.150 M H3PO4 is combined with 95.0 mL of 0.150 M NaOH, how many
grams of water are theoretically produced by the reaction?
H3PO4(aq) + 3NaOH(aq) → Na3PO4 (aq) + 3H2O (l)
35.0 mL 95.0 mL ? G
0.15M 0.150 M
1. Organize information
2. Which reactant do we use to find g of water?
0.0350 L H3PO4 x (0.15 mol / L) x (3 mol H2O / 1 mol H3PO4) x (18.016 g / 1 mol
H2O)
= 0.284 g H2O
*This above answer only has a 50% chance of being correct…so we have to do it
for NaOH too…
0.950 L NaOH x (0.15 mol / L) x (3 mol H2O / 3 mol NaOH) x (18.016 g / 1 mol
H2O)
= 0.257 g H2O
The second one is the answer because it is smaller, meaning that we are going
to run out of NaOH first, because it is the limiting reactant.
Solution Stoichiometry Practice
One way the US Environmental Protection Agency (EPA) tests for chloride contaminants
in water is by titrating a sample of silver nitrate solution. Any chloride anions in solution
will combine with the silver cations to produce bright white silver chloride precipitate.
Suppose an EPA chemist tests a 250. mL sample of groundwater know to be
contaminated with copper (II) chloride, which would react with silver nitrate solution like
this:
CuCl2 (aq) + 2AgNO3(aq) → 2AgCl(s) + Cu(NO3)2 (aq)
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Document Summary

H3po4(aq) + 3naoh(aq) na3po4 (aq) + 3h2o (l) 0. 0350 l h3po4 x (0. 15 mol / l) x (3 mol h2o / 1 mol h3po4) x (18. 016 g / 1 mol. *this above answer only has a 50% chance of being correct so we have to do it for naoh too . 0. 950 l naoh x (0. 15 mol / l) x (3 mol h2o / 3 mol naoh) x (18. 016 g / 1 mol. The second one is the answer because it is smaller, meaning that we are going to run out of naoh first, because it is the limiting reactant. One way the us environmental protection agency (epa) tests for chloride contaminants in water is by titrating a sample of silver nitrate solution. Any chloride anions in solution will combine with the silver cations to produce bright white silver chloride precipitate.

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