what mass of agcl will precipitate when 10.0 g of agno3 is added to an aqueous solution of nacl?

seven.v: Solution Stoichiometry

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Equally nosotros learned in Chapter 5, double replacement reactions involve the reaction between ionic compounds in solution and, in the course of the reaction, the ions in the 2 reacting compounds are "switched" (they replace each other). As an example, silver nitrate and sodium chloride react to grade sodium nitrate and the insoluble chemical compound, silver chloride.

\[\ce{AgNO3 (aq) + NaCl (aq) → AgCl (s) + NaNO3 (aq)}\]

Because these reactions occur in aqueous solution, we can use the concept of molarity to straight calculate the number of moles of products that will be formed, and hence the mass of precipitates. In the reaction shown above, if we mixed 123 mL of a one.00 Yard solution of NaCl with 72.five mL of a 2.71 M solution of AgNO ₃ , we could summate the moles (and hence, the mass) of AgCl that will be formed as follows:

First, we must examine the reaction stoichiometry. In this reaction, one mole of AgNO _iii reacts with one mole of NaCl to give one mole of AgCl. Because our ratios are one, we don't demand to include them in the equation. Next, we demand to summate the number of moles of each reactant:

\[0.123L\times \left ( \frac{1.00\: mole}{i.00\: Fifty} \right )=0.123\: moles\: NaCl\]

\[0.0725L\times \left ( \frac{2.71\: mole}{ane.00\: L} \right )=0.196\: moles\: AgNO_{3}\]

Because this is a limiting reactant problem, we need to recall that the moles of product that can be formed will equal the smaller of the number of moles of the 2 reactants. In this case, NaCl is limiting and AgNO ₃ is in backlog. Because our stoichiometry is one-to-one, we will therefore form 0.123 moles of AgCl. Finally, nosotros can catechumen this to mass using the molar mass of AgCl:

\[0.0725L\times \left ( \frac{ii.71\: mole}{1.00\: L} \right )=0.196\: moles\: AgNO_{3}\]

In a reaction where the stoichiometry is non one-to-i, you only need to include the stoichiometric ratio in you equations. Thus, for the reaction between lead (2) nitrate and potassium iodide, two moles of potassium iodide are required for every mole of pb (Ii) iodide that is formed.

\[\ce{Atomic number 82(NO3)two (aq) + 2 KI (aq) → PbI2 (s) + 2 KNO3 (aq)}\] For example: 1.78 grams of atomic number 82 (II) nitrate are dissolved in 17.0 mL of water and so mixed with 25.0 mL of 2.5 Thou potassium iodide solution. What mass of lead (II) iodide will be formed and what will be the final concentration of potassium nitrate in the solution? Again, we need to wait at this as a limiting reactant problem and outset calculate the number of moles of each reactant: \[1.78\: m\times \left ( \frac{1.00\: mole}{331.two\: g} \right )=five.37\times x^{-iii}\: moles\: Pb(NO_{3})_{2}\] \[0.0025\: Fifty\times \left ( \frac{2.50\: mole}{ane.00\: L} \correct )=six.25\times 10^{-3}\: moles\: KI\] The stoichiometry of this reaction is given past the ratios: \[\left ( \frac{1\: mole\: PbI_{ii}}{2\: mole\: KI} \right )\; and\; \left ( \frac{1\: mole\: PbI_{2}}{one\: mole\: Pb(NO_{iii})_{2}} \right )\] so the number of moles of production that would be formed from each reactant is calculated equally:

\[\left ( \frac{ane\: mole\: PbI_{2}}{ane\: mole\: Pb(NO_{3})_{ii}} \right )\]

\[half-dozen.25\times ten^{-3}\: moles\: KI\times \left ( \frac{i\: mole\: PbI_{2}}{2\: moles\: KI} \right )=3.12\times ten^{-3}\: moles\: PbI_{ii}\]

Potassium iodide produces the smaller amount of PbI ₂ and hence, is limiting and pb (II) nitrate is in excess. The mass of atomic number 82 (Ii) iodide that will exist produced is and so calculated from the number of moles and the molar mass:

\[3.12\times 10^{-3}\: moles\: \times \left ( \frac{461\: grams}{i\: mole} \right )=ane.44\: grams\: PbI_{ii}\]

To determine the concentration of potassium nitrate in the terminal solution, nosotros need to notation that ii moles of potassium nitrate are formed for every mole of PbI ₂ , or a stoichiometric ratio of \[\left ( \frac{2\: moles\: KNO_{3}}{1\: mole\: PbI_{2}} \right )\]

Our last volume is (17.0 + 25.0) = 42.0 mL, and the concentration of potassium nitrate is calculated as:

\[\frac{3.12\times x^{-3}\: moles\:PbI_{2}\times \left ( \frac{2\: moles\: KNO_{3}}{1\: mole\: PbI_{2}} \right )}{0.0420\: L}=0.148\; moles\; KNO_{3}/50\; or\; 0.148\; M\]

Exercise \(\PageIndex{ane}\)

1. A sample of 12.7 grams of sodium sulfate (Na ₂ SO ₄ ) is dissolved in 672 mL of distilled h2o.
   1. What is the molar concentration of sodium sulfate in the solution?
   2. What is the concentration of sodium ion in the solution?
2. How many moles of sodium sulfate must be added to an aqueous solution that contains 2.0 moles of barium chloride in order to precipitate 0.l moles of barium sulfate?
3. If one.0 g of NaN ₃ reacts with 25 mL of 0.20 One thousand NaNO ₃ according to the reaction shown below, how many moles of N ₂ (grand) are produced?

\[5 NaN3(s) + NaNO3(aq) → iii Na2O(due south) + 8 N2(g)\]

• Paul R. Young, Professor of Chemistry, University of Illinois at Chicago, Wiki: AskTheNerd; PRYaskthenerd.com - pyounguic.edu; ChemistryOnline.com

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Source: https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Book%3A_Introductory_Chemistry_Online_(Young)/07%3A_Aqueous_Solutions/7.5%3A_Solution_Stoichiometry

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