Magnesium is obtained from sea water. \(\mathrm{Ca}(\mathrm{OH})_{2}\) is added to sea water to precipitate \(\mathrm{Mg}(\mathrm{OH})_{2} .\) The precipitate is filtered and reacted with HCl to produce \(\mathrm{MgCl}_{2} .\) The \(\mathrm{MgCl}_{2}\) is electrolyzed to produce Mg and \(\mathrm{Cl}_{2} .\) If 185.0 \(\mathrm{g}\) of magnesium are recovered from \(1000 . \mathrm{gMgCl}_{2},\) what is the percentage yield for this reaction?

Short Answer

Expert verified
Percentage yield of magnesium is 72.50%.

Step by step solution

01

Write the relevant chemical reaction

The extraction process involving the electrolysis of magnesium chloride can be represented by the balanced chemical equation: \[ \text{MgCl}_{2} \rightarrow \text{Mg} + \text{Cl}_{2} \]
02

Determine the molar masses

Calculate the molar masses of \( \text{Mg} \) and \( \text{MgCl}_{2} \):Molar mass of \( \text{Mg} = 24.305 \text{ g/mol} \)Molar mass of \( \text{MgCl}_{2} = (24.305 + 2 \times 35.453) \text{ g/mol} = 95.211 \text{ g/mol} \)
03

Calculate the theoretical yield

First, determine the moles of \( \text{MgCl}_{2} \) in 1000 g:\[ \frac{1000 \text{ g}}{95.211 \text{ g/mol}} \text{ MgCl}_{2} = 10.50 \text{ mol} \]Using the stoichiometry of the reaction, 1 mole of \( \text{MgCl}_{2} \) produces 1 mole of \( \text{Mg} \).Thus, the theoretical yield in moles of \( \text{Mg} \) is 10.50 mol.Convert this to grams:\[ 10.50 \text{ mol} \times 24.305 \text{ g/mol} = 255.20 \text{ g} \]
04

Calculate the percentage yield

Percentage yield is calculated using the formula:\[ \text{Percentage yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100 \]Substituting the given values:\[ \text{Percentage yield} = \frac{185.0 \text{ g}}{255.20 \text{ g}} \times 100 \ = 72.50\text{ \text{ %} }\]

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Key Concepts

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

Chemical Reactions
Chemical reactions involve the transformation of reactants into products.
In the given exercise, several chemical reactions occur.
Seawater magnesium reacts with calcium hydroxide to precipitate magnesium hydroxide (Mg(OH)₂).
When hydrochloric acid (HCl) is added, it converts magnesium hydroxide to magnesium chloride (MgCl₂).
The chemical equation for the electrolysis of MgCl₂ is: \[ \text{MgCl}_{2} \rightarrow \text{Mg} + \text{Cl}_{2} \]
Electrolysis splits MgCl₂ into magnesium and chlorine gas.
Molar Mass
The molar mass of a compound is the mass of one mole of that substance.
It's calculated by adding up the atomic masses of all the atoms in the compound.
For example, to find the molar mass of magnesium chloride (MgCl₂), we add:
- Magnesium (Mg): 24.305 g/mol
- Chlorine (Cl): 2 x 35.453 g/mol
This gives a total molar mass for MgCl₂ of 95.211 g/mol.
This is crucial for converting between grams and moles in stoichiometric calculations.
Stoichiometry
Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction.
It allows us to predict the amounts of substances consumed and produced.
In the exercise, we calculated the moles of MgCl₂ from a given mass (1000 g) using its molar mass: \ \[\frac{1000 \text{ g}}{95.211 \text{ g/mol}} = 10.50 \text{ mol} \]
According to the balanced equation, 1 mole of MgCl₂ produces 1 mole of Mg.
Thus, 10.50 moles of MgCl₂ will produce 10.50 moles of Mg.
Electrolysis
Electrolysis is the process of using electricity to break down a compound into its elements.
In this exercise, electrolysis is used to extract magnesium (Mg) from magnesium chloride (MgCl₂).
The overall reaction for the electrolysis is: \[\text{MgCl}_{2} \rightarrow \text{Mg} + \text{Cl}_{2}\]
During electrolysis, Mg²⁺ ions gain electrons (reduction) to form Mg atoms.
The Cl⁻ ions lose electrons (oxidation) to form chlorine gas (Cl₂).
Yield Calculation
Yield calculation helps us determine the efficiency of a chemical reaction.
The theoretical yield is the maximum amount of product that can be formed based on stoichiometric calculations.
In the given exercise, the theoretical yield of magnesium is calculated from the initial amount of MgCl₂:
- Theoretical yield: 255.20 g
The actual yield is the amount of product actually obtained from the reaction.
Here, 185.0 g of Mg is recovered.
The percentage yield is calculated using:
\[ \text{Percentage yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100 = \frac{185.0 \text{ g}}{255.20 \text{ g}} \times 100 = 72.50\text{ %}\]
This shows how efficient the reaction was in producing the desired product.

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Most popular questions from this chapter

As early as \(1938,\) the use of NaOH was suggested as a means of removing \(\mathrm{CO}_{2}\) from the cabin of a spacecraft according to the following (unbalanced) reaction: \(\mathrm{NaOH}+\mathrm{CO}_{2} \longrightarrow \mathrm{Na}_{2} \mathrm{CO}_{3}+\mathrm{H}_{2} \mathrm{O}\) . a. If the average human body discharges 925.0 \(\mathrm{gCO}_{2}\) per day, how many moles of NaOH are needed each day for each person in the spacecraft? b. How many moles of each product are formed?

Analyzing Results Very seldom are chemists able to achieve a 100% yield of a product from a chemical reaction. However, the yield of a reaction is usually important because of the expense involved in producing less product. For example, when magnesium metal is heated in a crucible at high temperatures, the product magnesium oxide, MgO, is formed. Based on your analysis of the reaction, describe some of the actions that you would take to increase your percentage yield. The reaction is as follows: $$ 2 \mathrm{Mg}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{MgO}(s) $$

Calculate the indicated quantity for each of the various chemical reactions given: a. theoretical yield \(=20.0 \mathrm{g},\) actual yield \(=15.0 \mathrm{g}\) percentage yield \(=?\) b. theoretical yield \(=1.0 \mathrm{g},\) percentage yield \(=90.0 \%\) actual yield \(=?\) c. theoretical yield \(=5.00 \mathrm{g},\) actual yield \(=4.75 \mathrm{g}\) percentage yield \(=?\) d. theoretical yield \(=3.45 \mathrm{g}\) , percentage yield = 48.0\(\%\) , actual yield =?

Given the chemical equation \(\mathrm{Na}_{2} \mathrm{CO}_{3}(a q)+\mathrm{Ca}(\mathrm{OH})_{2}\) \(\longrightarrow 2 \mathrm{NaOH}(a q)+\mathrm{CaCO}_{3}(s),\) determine to two decimal places the molar masses of all substances involved. Then, write the molar masses as conversion factors.

a. Distinguish between the theoretical yield and actual yield in stoichiometric calculations. b. How does the value of the theoretical yield generally compare with the value of the actual yield?

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