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How many grams of chlorine gas can be produced from the electrolytic decomposition of \(874 \mathrm{~g}\) of calcium chloride? Name and calculate the mass (in \(\mathrm{g}\) ) of the other product.

Short Answer

Expert verified
279.26 g of chlorine gas, 315.67 g of calcium.

Step by step solution

01

Write the Balanced Chemical Equation

The decomposition reaction of calcium chloride (CaClâ‚‚) by electrolysis can be represented by the balanced chemical equation:\[ 2 \text{CaCl}_2 \rightarrow 2 \text{Ca} + \text{Cl}_2 \]
02

Molar Mass Calculation

Calculate the molar masses of the reactant and products:Molar mass of CaClâ‚‚: \[ 40.08 + 2 \times 35.45 = 110.98 \text{ g/mol} \]Molar mass of Clâ‚‚: \[ 2 \times 35.45 = 70.90 \text{ g/mol} \]Molar mass of Ca: \[ 40.08 \text{ g/mol} \]
03

Determine Moles of CaClâ‚‚

Convert the given mass of calcium chloride to moles:\[ \text{Moles of CaCl}_2 = \frac{874 \text{ g}}{110.98 \text{ g/mol}} = 7.875 \text{ mol} \]
04

Calculate Moles of Chlorine Gas Produced

According to the balanced equation, 2 moles of CaClâ‚‚ produce 1 mole of Clâ‚‚. Therefore, the moles of Clâ‚‚ produced from 7.875 moles of CaClâ‚‚ are:\[ \text{Moles of Cl}_2 = \frac{7.875 \text{ mol}}{2} = 3.9375 \text{ mol} \]
05

Calculate Mass of Chlorine Gas

Convert the moles of chlorine gas to mass:\[ \text{Mass of Cl}_2 = 3.9375 \text{ mol} \times 70.90 \text{ g/mol} = 279.26 \text{ g} \]
06

Calculate Mass of Other Product (Calcium)

According to the balanced equation, 2 moles of CaClâ‚‚ produce 2 moles of Ca. Therefore, the moles of Ca produced from 7.875 moles of CaClâ‚‚ are:\[ \text{Moles of Ca} = 7.875 \text{ mol} \]Convert the moles of calcium to mass:\[ \text{Mass of Ca} = 7.875 \text{ mol} \times 40.08 \text{ g/mol} = 315.67 \text{ g} \]

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Key Concepts

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

Balanced Chemical Equation
In chemistry, a balanced chemical equation is crucial for understanding the relationship between reactants and products in a chemical reaction. A chemical equation shows the substances involved in a reaction and their relative amounts. Balancing this equation ensures the law of conservation of mass is respected - meaning the number of atoms of each element is the same on both sides of the equation.

For example, in the electrolytic decomposition of calcium chloride (\text{CaCl\(_2\)}), the balanced equation is represented as: \[ 2 \text{CaCl}_2 \rightarrow 2 \text{Ca} + \text{Cl}_2 \] This tells us that 2 molecules of \text{CaCl\(_2\)} decompose to produce 2 atoms of calcium (\text{Ca}) and 1 molecule of chlorine gas (\text{Cl\(_2\)}).

It's important to ensure the equation is balanced before moving on to perform any further calculations. Balancing helps in predicting the amount of products formed from given reactants and vice versa. The coefficients in the balanced equation (like the '2' in front of \text{CaCl\(_2\)}) indicate the molar ratio of the reactants and products.
Molar Mass Calculation
Molar mass is used to convert between grams and moles, an essential step in stoichiometry. The molar mass of a compound is the total mass of all the atoms in one mole of that compound. It's calculated by adding up the atomic masses of all the atoms present in the molecular formula. For the decomposition of calcium chloride (\text{CaCl\(_2\)}), we calculate the molar mass as follows:

- Molar mass of \text{CaCl\(_2\)} is \[ 40.08 + 2 \times 35.45 = 110.98 \text{ g/mol} \] \text{Calcium} has a molar mass of 40.08 g/mol and \text{chlorine} has a molar mass of 35.45 g/mol.

Knowing the molar masses of \text{CaCl\(_2\)}, \text{Ca}, and \text{Cl\(_2\)} allows us to convert masses into moles, making it easier to use the stoichiometric ratios to determine how much of each product will be formed from a given amount of reactant. For instance, with \text{Cl\(_2\)}, the molar mass is \[ 2 \times 35.45 = 70.90 \text{ g/mol} \]
Stoichiometry
Stoichiometry is the calculation of reactants and products in chemical reactions. It involves using balanced chemical equations to determine the mole ratios of reactants and products. For the decomposition of calcium chloride, we use stoichiometric principles to find out how much chlorine gas (\text{Cl\(_2\)}) and calcium (\text{Ca}) can be produced from a given amount of calcium chloride (\text{CaCl\(_2\)}).

Steps in stoichiometric calculations:
  • Determine moles of \text{CaCl\(_2\)}: \[ \text{Moles of CaCl\(_2\)} = \frac{874 \text{ g}}{110.98 \text{ g/mol}} = 7.875 \text{ mol} \]
  • Find moles of \text{Cl\(_2\)} produced: According to the balanced equation, 2 moles of \text{CaCl\(_2\)} produce 1 mole of \text{Cl\(_2\)}. So, \[ \text{Moles of Cl\(_2\)} = \frac{7.875 \text{ mol}}{2} = 3.9375 \text{ mol} \]
  • Convert moles of \text{Cl\(_2\)} to grams: \[ \text{Mass of Cl\(_2\)} = 3.9375 \text{ mol} \times 70.90 \text{ g/mol} = 279.26 \text{ g} \]
  • For calcium, 2 moles of \text{CaCl\(_2\)} produce 2 moles of \text{Ca}. So, \[ \text{Moles of Ca} = 7.875 \text{ mol} \]
  • Then convert moles of \text{Ca} to grams: \[ \text{Mass of Ca} = 7.875 \text{ mol} \times 40.08 \text{ g/mol} = 315.67 \text{ g} \]
Using stoichiometry helps predict the quantities of products formed and ensure that no reactant is wasted. It aligns chemical calculations with the balanced equation, providing precise results.

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Most popular questions from this chapter

A sample of impure magnesium was analyzed by allowing it to react with excess HCl solution. After \(1.32 \mathrm{~g}\) of the impure metal was treated with \(0.100 \mathrm{~L}\) of \(0.750 \mathrm{M} \mathrm{HCl}, 0.0125 \mathrm{~mol}\) of \(\mathrm{HCl}\) remained. Assuming the impurities do not react, what is the mass \% of Mg in the sample?

If \(38.5 \mathrm{~mL}\) of lead(II) nitrate solution reacts completely with excess sodium iodide solution to yield \(0.628 \mathrm{~g}\) of precipitate, what is the molarity of lead(II) ion in the original solution?

Why must every redox reaction involve an oxidizing agent and a reducing agent?

A mathematical equation useful for dilution calculations is \(M_{\text {dil }} \times V_{\text {dil }}=M_{\text {cunc }} \times V_{\text {cvac }}\). (a) What does each symbol mean, and why does the equation work? (b) Given the volume and molarity of a \(\mathrm{CaCl}_{2}\) solution, how do you determine the amount (mol) and the mass (g) of solute?

A reaction cycle for an element is a series of reactions beginning and ending with that element. In the following copper reaction cycle, copper has either a 0 or a +2 oxidation state. Write balanced molecular and net ionic equations for each step. (1) Copper metal reacts with aqueous bromine to produce a greenblue solution. (2) Adding aqueous sodium hydroxide forms a blue precipitate. (3) The precipitate is heated and turns black (water is released). (4) The black solid dissolves in nitric acid to give a blue solution. (5) Adding aqueous sodium phosphate forms a green precipitate. (6) The precipitate forms a blue solution in sulfuric acid. (7) Copper metal is recovered from the blue solution when zinc metal is added.

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