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Chapter 11: Problem 33

Small quantities of chlorine can be prepared in the laboratory by the reaction described by the equation $$ \begin{aligned} \mathrm{MnO}_{2}(s)+4 \mathrm{HCl}(a q) \rightarrow & \\ & \mathrm{MnCl}_{2}(a q)+\mathrm{Cl}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) \end{aligned} $$ How many grams of chlorine can be prepared from \(100.0\) grams of manganese(II) oxide?

Short Answer

Expert verified
81.54 grams of chlorine can be prepared from 100.0 grams of manganese(II) oxide.

Step by step solution

01

Determine the Molar Masses

To calculate the grams of chlorine that can be prepared, the molar mass of each reactant and product must be determined.- Molar mass of MnO鈧�: Mn = 54.94 g/mol, O = 16.00 g/mol \[ \text{Molar mass of MnO}_2 = 54.94 + (2 \times 16.00) = 86.94 \text{ g/mol} \]- Molar mass of Cl鈧�: Cl = 35.45 g/mol \[ \text{Molar mass of Cl}_2 = 2 \times 35.45 = 70.90 \text{ g/mol} \]
02

Calculate Moles of Manganese(II) Oxide

Using the molar mass of MnO鈧�, calculate the number of moles in 100.0 grams of MnO鈧�.\[ \text{Moles of MnO}_2 = \frac{100.0 \text{ g}}{86.94 \text{ g/mol}} = 1.150 \text{ mol} \]
03

Apply Stoichiometry to Find Moles of Chlorine

According to the balanced chemical equation, 1 mole of MnO鈧� produces 1 mole of Cl鈧�. Thus, the moles of Cl鈧� produced is the same as the moles of MnO鈧�.\[ \text{Moles of Cl}_2 = 1.150 \text{ mol} \]
04

Convert Moles of Chlorine to Grams

Convert the moles of chlorine to grams using the molar mass.\[ \text{Grams of Cl}_2 = 1.150 \text{ mol} \times 70.90 \text{ g/mol} = 81.54 \text{ g} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Reactions
Chemical reactions occur when substances interact to form new substances with different properties. These interactions are described using chemical equations, which outline the reactants that start a reaction and the products formed by the end. In this example, manganese(II) oxide (MnO鈧�) reacts with hydrochloric acid (HCl) to form manganese(II) chloride (MnCl鈧�), chlorine gas (Cl鈧�), and water (H鈧侽). The balanced chemical equation ensures that the number of atoms of each element is the same on both sides, adhering to the law of conservation of mass.
Understanding the reaction equation is crucial because it shows the molar ratio of the reactants and products. This ratio is used to determine the amount of product that can be formed from a given amount of reactant. In our exercise, for every mole of MnO鈧� used, one mole of Cl鈧� is produced. This is an example of stoichiometry, allowing us to predict the outcome of a chemical reaction.
Molar Mass
The molar mass is the mass of one mole of a substance, typically measured in grams per mole (g/mol). It is essential for converting between the mass of a substance and the number of moles. Molar mass is determined by adding the atomic masses of all the atoms in a molecule.
For example, to find the molar mass of MnO鈧�, we add the atomic masses of manganese (54.94 g/mol) and oxygen (16.00 g/mol each) resulting in a molar mass of 86.94 g/mol. Similarly, the molar mass of Cl鈧� is calculated by adding the atomic mass of chlorine (35.45 g/mol) multiplied by two, giving us 70.90 g/mol.
Knowing the molar mass is crucial for calculations in chemistry, as it allows us to convert from grams, a practical measure used in the lab, to moles, which are used in theoretical calculations.
Moles Calculation
Calculating moles is a vital step in stoichiometry and is essential for determining the quantities involved in a chemical reaction. A mole is a unit that measures the amount of substance. It is similar to counting by dozens or hundreds and is defined by Avogadro's number, approximately 6.022 x 10虏鲁 representative particles per mole.
In our example, we start by calculating the moles of MnO鈧� using its molar mass. With 100.0 grams of MnO鈧� and a molar mass of 86.94 g/mol, we determine there are 1.150 moles of MnO鈧�. The balanced equation tells us this will produce an equal number of moles of chlorine gas, since the ratio is 1:1.
Finally, to find the mass of chlorine produced, we convert moles back to grams using the molar mass of Cl鈧�, arriving at 81.54 grams of chlorine. This conversion ensures that the theoretical stoichiometric calculations can be applied practically in the laboratory setting.

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Most popular questions from this chapter

11-62. Ethanol, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}(l)\), is produced commercially from the reaction of water with ethene, \(\mathrm{C}_{2} \mathrm{H}_{4}(g)\), commonly known as ethylene. Ethylene is obtained from petroleum and is one of the top produced industrial chemicals in the United States (Appendix \(\mathrm{H}\) ). It is the basis for the synthesis of a variety of important chemicals and polymers. The reaction equation for the synthesis of ethanol is $$ \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(l) \stackrel{\mathrm{H}_{2} \mathrm{SO}_{4}}{\longrightarrow} \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}(l) $$ Given that \(13.5\) kilograms of ethanol were produced from \(10.0\) kilograms of ethylene, calculate the percentage yield in this synthesis.

II-53. Cryolite, \(\mathrm{Na}_{3} \mathrm{AlF}_{6}(s)\), an ore used in the production of aluminum, can be synthesized by the reaction described by the equation \(\mathrm{Al}_{2} \mathrm{O}_{3}(s)+\mathrm{NaOH}(l)+\mathrm{HF}(g) \stackrel{\text { High } T}{\longrightarrow}\) \(\mathrm{Na}_{3} \mathrm{AlF}_{6}(s)+\mathrm{H}_{2} \mathrm{O}(g) \quad(\) unbalanced \()\) (a) Balance this equation. (b) If \(10.0\) kilograms of \(\mathrm{Al}_{2} \mathrm{O}_{3}(s), 50.00\) kilograms of \(\mathrm{NaOH}(l)\), and \(50.0\) kilograms of \(\mathrm{HF}(g)\) react completely, how many kilograms of cryolite will be produced? (c) Which reactants will be in excess and how many kilograms of each of these reactants will remain?

MOLECULAR FORMULAS The chemical 2-propanone, commonly known as acetone, is an important chemical solvent; a familiar home use is as a nail polish remover. Chemical analysis shows that acetone is \(62.0 \%\) carbon, \(10.4 \%\) hydrogen, and \(27.5 \%\) oxygen by mass. Determine the empirical formula of acetone. In a separate experiment, the molecular mass is found to be \(58.1 .\) What is the molecular formula of acetone?

Why is it not possible for a reaction to have a percentage yield of the desired product greater than \(100 \%\) (what law is this in violation of)? What is the most likely explanation for a percentage yield measured in the laboratory that is in excess of \(100 \%\) ?

Bromine can be prepared by adding chlorine to an aqueous solution of sodium bromide. The reaction equation is $$ 2 \mathrm{NaBr}(a q)+\mathrm{Cl}_{2}(g) \rightarrow \mathrm{Br}_{2}(l)+2 \mathrm{NaCl}(a q) $$ How many grams of bromine are formed if a solution containing \(25.0\) grams of aqueous \(\operatorname{NaBr}(s)\) and \(25.0\) grams of \(\mathrm{Cl}_{2}(g)\) are reacted?
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