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Chapter 4: Problem 115

Calculate the mass of the precipitate formed when \(2.27 \mathrm{~L}\) of \(0.0820 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) are mixed with \(3.06 \mathrm{~L}\) of \(0.0664 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}\).

Short Answer

Expert verified
The mass of the precipitate formed is 43.43 g.

Step by step solution

01

Write down the balanced chemical equation

The balanced chemical equation for the reaction is: \[Ba(OH)_2 + Na_2SO_4 \rightarrow BaSO_4 \downarrow + 2NaOH\]. The arrow pointing downwards (\(\downarrow\)) after BaSO_4 indicates that it forms a precipitate.
02

Calculate moles of each compound

The number of moles of each compound can be calculated using molarity (M) and volume (V), with the formula n = MV (where n is the number of moles, M is the molarity and V is the volume). For Ba(OH)2: \[n = MV = 0.0820 \mathrm{mol/L} \times 2.27 \mathrm{L} = 0.18614 \mathrm{mol}\] For Na2SO4: \[n = MV = 0.0664 \mathrm{mol/L} \times 3.06 \mathrm{L} = 0.20304 \mathrm{mol}\]
03

Identify the limiting reactant

The stoichiometry of the balanced chemical equation shows us that the reaction requires one mole of barium hydroxide to react with one mole of sodium sulfate. As both reactants are not present in equal moles, one will limit the reaction. Comparing the number of moles of each compound, it's clear that Ba(OH)2 is the limiting reactant because we have less of it (0.18614 moles < 0.20304 moles). Hence, the amount of precipitate formed will be based on the amount of Ba(OH)2 present.
04

Calculate the mass of BaSO4 precipitate

The mass of the barium sulfate precipitate can be calculated from the number of moles of the limiting reactant \(Ba(OH)_2\), and using the molar mass of \(BaSO_4\). One mole of \(Ba(OH)_2\) produces one mole of \(BaSO_4\), hence the number of moles of \(BaSO_4\), will be the same as that of the limiting reactant \(Ba(OH)_2\). So, the mass of the precipitate is given by: \(m = n \times M_w\), where \(M_w\) is the molar mass of \(BaSO_4\) and is calculated to be \(233.39g/mol\). So the mass of \(BaSO_4\) precipitate is \(0.18614 mol \times 233.39 g/mol = 43.43 g\)

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Most popular questions from this chapter

Oxalic acid \(\left(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\right)\) is present in many plants and vegetables. If \(24.0 \mathrm{~mL}\) of \(0.0100 \mathrm{M} \mathrm{KMnO}_{4}\) solution is needed to titrate \(1.00 \mathrm{~g}\) of a sample of \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) to the equivalence point, what is the percent by mass of \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) in the sample? The net ionic equation is \(2 \mathrm{MnO}_{4}^{-}+16 \mathrm{H}^{+}+5 \mathrm{C}_{2} \mathrm{O}_{4}^{2-} \longrightarrow\) \(2 \mathrm{Mn}^{2+}+10 \mathrm{CO}_{2}+8 \mathrm{H}_{2} \mathrm{O}\)

The molar mass of a certain metal carbonate, \(\mathrm{MCO}_{3}\), can be determined by adding an excess of \(\mathrm{HCl}\) acid to react with all the carbonate and then "back titrating" the remaining acid with a \(\mathrm{NaOH}\) solution. (a) Write an equation for these reactions. (b) In a certain experiment, \(20.00 \mathrm{~mL}\) of \(0.0800 \mathrm{M} \mathrm{HCl}\) were added to a 0.1022 -g sample of \(\mathrm{MCO}_{3}\). The excess HCl required \(5.64 \mathrm{~mL}\) of \(0.1000 \mathrm{M} \mathrm{NaOH}\) for neutralization. Calculate the molar mass of the carbonate and identify \(\mathrm{M}\).

Write the equation that enables us to calculate the concentration of a diluted solution. Give units for all the terms.

Calculate the volume in milliliters of a \(1.420 \mathrm{M}\) \(\mathrm{NaOH}\) solution required to titrate the following solutions. (a) \(25.00 \mathrm{~mL}\) of a \(2.430 \mathrm{M} \mathrm{HCl}\) solution (b) \(25.00 \mathrm{~mL}\) of a \(4.500 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) solution (c) \(25.00 \mathrm{~mL}\) of a \(1.500 \mathrm{M} \mathrm{H}_{3} \mathrm{PO}_{4}\) solution

A student carried out two titrations using a \(\mathrm{NaOH}\) solution of unknown concentration in the buret. In one titration she weighed out \(0.2458 \mathrm{~g}\) of KHP (see Section 4.7 ) and transferred it to an Erlenmeyer flask. She then added \(20.00 \mathrm{~mL}\) of distilled water to dissolve the acid. In the other titration she weighed out \(0.2507 \mathrm{~g}\) of KHP but added \(40.00 \mathrm{~mL}\) of distilled water to dissolve the acid. Assuming no experimental error, would she obtain the same result for the concentration of the \(\mathrm{NaOH}\) solution?
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