A 2.50 g sample of \(\mathrm{Ag}_{2} \mathrm{SO}_{4}(\mathrm{s})\) is added to a beaker containing 0.150 L of 0.025 M BaCl\(_2\) (a) Write an equation for any reaction that occurs. (b) Describe the final contents of the beaker- -that is, the masses of any precipitates present and the concentrations of the ions in solution.

Short Answer

Expert verified
The balanced chemical reaction is \(Ag_2SO_4 + BaCl_2 \rightarrow 2AgCl (\downarrow) + BaSO_4 (\downarrow)\). The precipitates formed are AgCl and BaSO4 with masses 1.074 g and 0.875 g respectively. The Ag and SO4 ions have a concentration of 0 M in the solution while Ba and Cl ions are present in the excess Ag2SO4.

Step by step solution

01

Write the Balanced Chemical Reaction

When Ag2SO4 and BaCl2 react, they undergo a double replacement reaction. In this reaction, the silver (Ag) ions react with the chloride (Cl) ions, and the barium (Ba) ions react with the sulfate (SO4) ions. This reaction can be written as follows: \[ Ag_2SO_4 + BaCl_2 \rightarrow 2AgCl (\downarrow) + BaSO_4 (\downarrow) \] (^) The down arrow indicates that the product is a precipitate. The equation is already balanced, so it's not necessary to add any coefficients.
02

Identify the Limiting Reagent

This step involves finding the limiting reagent which will determine how much precipitate (product) can be formed. The number of moles of each reactant can be calculated using the given quantities. The number of moles of Ag2SO4 is \(\frac{2.50 g}{331.8 g/mol} = 0.0075 mol\). Since the BaCl2 solution is 0.025 M and has a volume of 0.150 L, the moles of BaCl2 can be calculated as \((0.025 mol/L) \times 0.150 L = 0.00375 mol\). As the reaction ratio of Ag2SO4 to BaCl2 is 1:1, it's clear that BaCl2 is the limiting reagent because there are fewer moles of it.
03

Compute the Mass and Concentration of the Precipitates

Using the molar ratios from the chemical equation, the amount of precipitates formed can be computed. For every mole of BaCl2, 2 moles of AgCl and 1 mole of BaSO4 are produced. Therefore, given that there are 0.00375 moles of BaCl2, there will be 0.00375 moles of BaSO4 and 0.0075 moles of AgCl produced. The masses of these substances can be computed by multiplying by the respective molar mass: For BaSO4, 0.00375 moles * (233.4 g/mol) = 0.875 g; and for AgCl, 0.0075 moles * (143.3 g/mol) = 1.074 g. The concentration of the ions in solution should be 0 for Ag and SO4 ions as all have reacted to form precipitates. Ba and Cl ions only remain in the excess reagent Ag2SO4, with concentration equal to (initial concentration - consumed concentration).
04

Describe the Solution and Final Contents of the Beaker

From the computed data, the final content of the beaker is as follow: the precipitates are AgCl and BaSO4 with masses 1.074 g and 0.875 g respectively, the concentrations of the Ag and SO4 ions in the solution are 0 M, and the concentrations of the Ba and Cl ions are present in the excess Ag2SO4.

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Most popular questions from this chapter

The following \(K_{\mathrm{sp}}\) values are found in a handbook. Write the solubility product expression to which each one applies. For example, \(K_{\mathrm{sp}}(\mathrm{AgCl})=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right]=\) \(1.8 \times 10^{-10}\). (a) \(K_{\mathrm{sp}}\left(\mathrm{Cr} \mathrm{F}_{3}\right)=6.6 \times 10^{-11}\) (b) \(K_{\mathrm{sp}}\left[\mathrm{Au}_{2}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\right]=1 \times 10^{-10}\) (c) \(K_{\mathrm{sp}}\left[\mathrm{Cd}_{3}\left(\mathrm{PO}_{4}\right)_{2}\right]=2.1 \times 10^{-33}\) (d) \(K_{\mathrm{sp}}\left(\mathrm{Sr} \mathrm{F}_{2}\right)=2.5 \times 10^{-9}\)

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