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Chapter 4: Q86E (page 224)

Sodium bicarbonate (baking soda), NaHCO3, can be purified by dissolving it in hot water (60掳C), filtering to remove insoluble impurities, cooling to 0 掳C to precipitate solid NaHCO3 , and then filtering to remove the solid, leaving soluble impurities in solution. Any NaHCO3 that remains in solution is not recovered. The solubility of NaHCO3 in hot water of 60 掳C is 164 g/L. Its solubility in cold water of 0 掳C is 69 g/L. What is the percent yield of NaHCO3 when it is purified by this method?

Short Answer

Expert verified

The percent yield of sodium bicarbonate when purified by the given will be 58%

Step by step solution

01

Given Data

It is given that the solubility of sodium bicarbonate in hot water is 164 g/L which is the theoretical yield of \({\rm{NaHC}}{{\rm{O}}_3}\)

Now, we know that solubility means the solute is dissolved in the solution, and in cold water the \({\rm{NaHC}}{{\rm{O}}_3}\) is present along with the other impurities, so the amount of \({\rm{NaHC}}{{\rm{O}}_3}\) that is present in the solution or the actual yield of \({\rm{NaHC}}{{\rm{O}}_3}\) will be 164 鈥� 69 = 95 g/L

02

Percent yield of sodium bicarbonate

The net percentage yield of the compound is calculated by:

\({\rm{Percentage yield = }}\frac{{{\rm{Actual yield}}}}{{{\rm{Theoretical yield}}}}{\rm{ }} \times {\rm{ 100}}\)

Therefore, the percentage yield of the \({\rm{NaHC}}{{\rm{O}}_3}\) is

\({\rm{Percentage yield = }}\frac{{95}}{{164}}{\rm{ }} \times {\rm{ 100 = 58 percent}}\)

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Most popular questions from this chapter

What is the percent of chloride ion in a sample if 1.324g of sample produces 1.0881g of AgCl when treated with excess Ag+\(A{g^ + } + C{l^ - } \to AgCl\)

Use the following equations to answer the next four questions:

i.\({H_2}O\left( s \right) \to {H_2}O\left( l \right)\)

ii.\(N{a^ + }\left( {aq} \right) + C{l^ - }\left( {aq} \right) + A{g^ + }\left( {aq} \right) + NO_3^ - \left( {aq} \right) \to AgCl\left( s \right) + N{a^ + }\left( {aq} \right) + NO_3^ - \left( {aq} \right)\)

iii.\(C{H_3}OH\left( g \right) + {O_2}\left( g \right) \to C{O_2}\left( g \right) + {H_2}O\left( g \right)\)

iv. \(2{H_2}O\left( l \right) \to 2{H_2}\left( g \right) + {O_2}\left( g \right)\)

v. \({H^ + }\left( {aq} \right) + O{H^ - }\left( {aq} \right) \to {H_2}O\left( l \right)\)

(a) Which equation describes a physical change?

(b)Which equation identifies the reactants and products of a combustion reaction?

(c)Which equation is not balanced?

(d)Which is a net ionic equation?

Potassium acid phthalate, KHC6H4O4or KHP, is used in many laboratories, including general chemistry laboratories, to standardize solutions of base. KHP is one of only a few stable solid acids that can be dried by warming and weighed. A 0.3420-g sample of KHC6H4O4 reacts with 35.73 mL of a NaOH solution in a titration. What is the molar concentration of the NaOH?

\({\rm{KH}}{{\rm{C}}_6}{{\rm{H}}_4}{{\rm{O}}_{4({\rm{aq}})}}{\rm{ + NaO}}{{\rm{H}}_{({\rm{aq}})}}{\rm{ }} \to {\rm{ KNa}}{{\rm{C}}_6}{{\rm{H}}_4}{{\rm{O}}_{4({\rm{aq}})}}{\rm{ + }}{{\rm{H}}_2}{{\rm{O}}_{({\rm{aq}})}}\)

Uranium can be isolated from its ores by dissolving it as UO2(NO3)2, then separating it as solid UO2(C2O4). Addition of 0.4031 g of sodium oxalate, NaC2O4, to a solution containing 1.481 g of uranyl nitrate, UO2(NO3)2, yields 1.073 g of solid

\(Na{C_2}{O_4} + U{O_2}{\left( {N{O_3}} \right)_2} + 3{H_2}O \to U{O_2}\left( {{C_2}{O_4}} \right) \cdot 3{H_2}O + 2NaN{O_3}\)

Titration of a 20.0-mL sample of acid rain required 1.7 mL of 0.0811MNaOH to reach the end point. If we assume that the acidity of the rain is due to the presence of sulfuric acid, what was the concentration of sulfuric acid in this sample of rain?
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