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Chapter 4: Problem 114

Some of the substances commonly used in stomach antacids are \(\mathrm{MgO}, \mathrm{Mg}(\mathrm{OH})_{2},\) and \(\mathrm{Al}(\mathrm{OH})_{3.}\) a. Write a balanced equation for the neutralization of hydrochloric acid by each of these substances. b. Which of these substances will neutralize the greatest amount of 0.10 M HCl per gram?

Short Answer

Expert verified
The balanced equations for the neutralization of hydrochloric acid by MgO, Mg(OH)鈧�, and Al(OH)鈧� are: a. \(MgO + 2HCl \to MgCl_{2} + H_{2}O\) b. \(Mg(OH)_{2} + 2HCl \to MgCl_{2} + 2H_{2}O\) c. \(Al(OH)_{3} + 3HCl \to AlCl_{3} + 3H_{2}O\) Moles of HCl neutralized per gram: - MgO: 0.0496 moles HCl/g - Mg(OH)鈧�: 0.0343 moles HCl/g - Al(OH)鈧�: 0.0385 moles HCl/g The substance that neutralizes the greatest amount of HCl per gram is MgO, with 0.0496 moles HCl/g.

Step by step solution

01

Write balanced equations for the neutralization

Let's write balanced equations for the neutralization of hydrochloric acid by MgO, Mg(OH)鈧�, and Al(OH)鈧�. a. For MgO: \(MgO + 2HCl \to MgCl_{2} + H_{2}O\) b. For Mg(OH)鈧�: \(Mg(OH)_{2} + 2HCl \to MgCl_{2} + 2H_{2}O\) c. For Al(OH)鈧�: \(Al(OH)_{3} + 3HCl \to AlCl_{3} + 3H_{2}O\)
02

Calculate moles of HCl neutralized per gram

To find which of these substances can neutralize the greatest amount of 0.10 M HCl per gram, we'll first need to calculate the number of moles of HCl neutralized per gram of each substance. a. For MgO: Molar mass of MgO: Mg (24.31 g/mol) + O (16.00 g/mol) = 40.31 g/mol Moles of HCl neutralized per gram of MgO: (1 mole MgO / 40.31 g) * 2 moles HCl / 1 mole MgO = 0.0496 moles HCl/g b. For Mg(OH)鈧�: Molar Mass of Mg(OH)鈧�: Mg (24.31 g/mol) + 2 * O (16.00 g/mol) + 2 * H (1.01 g/mol) = 58.33 g/mol Moles of HCl neutralized per gram of Mg(OH)鈧�: (1 mole Mg(OH)鈧� / 58.33 g) * 2 moles HCl / 1 mole Mg(OH)鈧� = 0.0343 moles HCl/g c. For Al(OH)鈧�: Molar Mass of Al(OH)鈧�: Al (26.98 g/mol) + 3 * O (16.00 g/mol) + 3 * H (1.01 g/mol) = 78.01 g/mol Moles of HCl neutralized per gram of Al(OH)鈧�: (1 mole Al(OH)鈧� / 78.01 g) * 3 moles HCl / 1 mole Al(OH)鈧� = 0.0385 moles HCl/g
03

Determine the substance that neutralizes the greatest amount of HCl per gram

Now that we have the moles of HCl neutralized per gram for each substance, we can compare them: MgO: 0.0496 moles HCl/g Mg(OH)鈧�: 0.0343 moles HCl/g Al(OH)鈧�: 0.0385 moles HCl/g The substance that neutralizes the greatest amount of HCl per gram is MgO, with 0.0496 moles HCl/g.

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Key Concepts

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

Balanced Chemical Equations
A balanced chemical equation is essential for accurately representing a chemical reaction. It shows the exact ratio in which reactants combine to form products. Each chemical equation is like a recipe, ensuring we start with the right amounts to get the desired outcome.

This involves matching the number of atoms for each element on both sides of the equation. For instance, in the neutralization of hydrochloric acid (HCl) with magnesium oxide (MgO), the equation is: \(MgO + 2HCl \to MgCl_2 + H_2O\). Here, we see that one MgO molecule reacts with two HCl molecules.

In another example, neutralizing with magnesium hydroxide (Mg(OH)鈧�) requires the equation: \(Mg(OH)_2 + 2HCl \to MgCl_2 + 2H_2O\), again noting the two HCl molecules.

Lastly, aluminium hydroxide (Al(OH)鈧�) follows this balanced equation: \(Al(OH)_3 + 3HCl \to AlCl_3 + 3H_2O\), showing it requires three HCl molecules. Each equation demonstrates how to balance the reactants and products properly.
Stoichiometry
Stoichiometry is the heart of quantitative chemistry. It's the process of calculating the amounts of reactants and products involved in a chemical reaction. Using a balanced chemical equation, stoichiometry helps us determine how much of each reactant is needed and how much product is produced.

This process relies on mole-to-mole conversion from the coefficients in the balanced equations. For instance, consider the equations in our antacid reactions. Each formula, like \(Mg(OH)_2 + 2HCl \to MgCl_2 + 2H_2O\), plainly illustrates that two moles of HCl react with one mole of Mg(OH)鈧�.

These ratios allow us to translate a chemical reaction into usable data, predicting how much acid can be neutralized. Stoichiometry is incredibly useful in laboratory and industrial applications, as it guides chemists in scaling up reactions efficiently and accurately.
Molar Mass Calculations
Molar mass calculations are critical to understanding chemical reactions and their outcomes. The molar mass is the weight of one mole of a given substance, expressed in grams per mole (g/mol). It combines the atomic masses of each atom in the molecule.

For example, the molar mass of magnesium oxide (MgO) is calculated by adding the masses of magnesium (24.31 g/mol) and oxygen (16.00 g/mol), resulting in 40.31 g/mol.

Similarly, the molar mass of magnesium hydroxide (Mg(OH)鈧�) is 58.33 g/mol. This involves summing the atomic masses of magnesium, two oxygen atoms, and two hydrogen atoms.

Aluminium hydroxide (Al(OH)鈧�) has a molar mass of 78.01 g/mol, derived from the masses of aluminium, three oxygen atoms, and three hydrogen atoms.

Knowing these molar masses is crucial, as they allow us to calculate how many moles are in a given mass and predict how much acid a certain weight of antacid can neutralize.

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Most popular questions from this chapter

Consider the reaction of 19.0 g of zinc with excess silver nitrite to produce silver metal and zinc nitrite. The reaction is stopped before all the zinc metal has reacted and 29.0 g of solid metal is present. Calculate the mass of each metal in the 29.0-g mixture

Consider a 1.50-g mixture of magnesium nitrate and magnesium chloride. After dissolving this mixture in water, 0.500 M silver nitrate is added dropwise until precipitate formation is complete. The mass of the white precipitate formed is 0.641 g. a. Calculate the mass percent of magnesium chloride in the mixture. b. Determine the minimum volume of silver nitrate that must have been added to ensure complete formation of the precipitate.

Specify which of the following equations represent oxidation鈥� reduction reactions, and indicate the oxidizing agent, the reducing agent, the species being oxidized, and the species being reduced a. \(\mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightarrow \mathrm{CO}(g)+3 \mathrm{H}_{2}(g)\) b. \(2 \mathrm{AgNO}_{3}(a q)+\mathrm{Cu}(s) \rightarrow \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(a q)+2 \mathrm{Ag}(s)\) c. \(\mathrm{Zn}(s)+2 \mathrm{HCl}(a q) \rightarrow \mathrm{ZnCl}_{2}(a q)+\mathrm{H}_{2}(g)\) d. \(2 \mathrm{H}^{+}(a q)+2 \mathrm{CrO}_{4}^{2-}(a q) \rightarrow \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\)

In most of its ionic compounds, cobalt is either Co(II) or Co(III). One such compound, containing chloride ion and waters of hydration, was analyzed, and the following results were obtained. A 0.256-g sample of the compound was dissolved in water, and excess silver nitrate was added. The silver chloride was filtered, dried, and weighed, and it had a mass of 0.308 g. A second sample of 0.416 g of the compound was dissolved in water, and an excess of sodium hydroxide was added. The hydroxide salt was filtered and heated in a flame, forming cobalt(III) oxide. The mass of cobalt(III) oxide formed was 0.145 g. a. What is the percent composition, by mass, of the compound? b. Assuming the compound contains one cobalt ion per formula unit, what is the formula? c. Write balanced equations for the three reactions described.

Many plants are poisonous because their stems and leaves contain oxalic acid, \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4},\) or sodium oxalate, \(\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) . When ingested, these substances cause swelling of the respiratory tract and suffocation. A standard analysis for determining the amount of oxalate ion, \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-},\) in a sample is to precipitate this species as calcium oxalate, which is insoluble in water. Write the net ionic equation for the reaction between sodium oxalate and calcium chloride, \(\mathrm{CaCl}_{2},\) in aqueous solution.
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