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Calculate the number of moles of: (a) \(\mathrm{S}^{2-}\) ions in \(6.56 \mathrm{mol}\) of \(\mathrm{Na}_{2} \mathrm{S}\) (b) \(\mathrm{Mg}^{2+}\) ions in \(8.320 \mathrm{mol}\) of \(\mathrm{Mg}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) (c) acetate ions, \(\mathrm{CH}_{3} \mathrm{COO}^{-},\) in \(0.43 \mathrm{mol}\) of \(\mathrm{Ca}\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{2}\)

Short Answer

Expert verified
(a) 6.56 mol of \( \mathrm{S}^{2-} \), (b) 24.96 mol of \( \mathrm{Mg}^{2+} \), (c) 0.86 mol of acetate ions.

Step by step solution

01

Analyze Moles of \( \mathrm{S}^{2-} \) Ions

Given the substance \( \mathrm{Na}_2 \mathrm{S} \), each formula unit contains one \( \mathrm{S}^{2-} \) ion. Therefore, the number of moles of \( \mathrm{S}^{2-} \) ions is equal to the number of moles of \( \mathrm{Na}_2 \mathrm{S} \). Hence, the number of moles of \( \mathrm{S}^{2-} \) ions is **6.56 mol**.
02

Analyze Moles of \( \mathrm{Mg}^{2+} \) Ions

The compound \( \mathrm{Mg}_3(\mathrm{PO}_4)_2 \) contains three \( \mathrm{Mg}^{2+} \) ions per formula unit. Therefore, to find the number of moles of \( \mathrm{Mg}^{2+} \) ions, multiply the moles of \( \mathrm{Mg}_3(\mathrm{PO}_4)_2 \) by 3. Thus, \( 8.320 \text{ mol} \times 3 = 24.96 \text{ mol} \) of \( \mathrm{Mg}^{2+} \) ions.
03

Analyze Moles of Acetate Ions

Each formula unit of \( \mathrm{Ca}(\mathrm{CH}_3\mathrm{COO})_2 \) contains two acetate ions \( (\mathrm{CH}_3 \mathrm{COO}^-) \). Therefore, multiply the moles of \( \mathrm{Ca}(\mathrm{CH}_3\mathrm{COO})_2 \) by 2 to find the moles of acetate ions: \( 0.43 \text{ mol} \times 2 = 0.86 \text{ mol} \) of \( \mathrm{CH}_3\mathrm{COO}^- \) ions.

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Key Concepts

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

Stoichiometry
Stoichiometry is an essential concept in chemistry that involves calculating the relationships between reactants and products in a chemical reaction. It provides a way to predict how substances interact and how much of each is needed or produced in a reaction.
To understand stoichiometry, it's crucial to recognize that chemical equations are balanced to ensure that the same number of atoms of each element is present on both the reactant and product sides. This balance maintains the principle of conservation of mass.
When solving stoichiometric problems, you'll often use the mole concept, which acts as a bridge between the atomic scale and the macroscopic scale.
**Steps to Follow in Stoichiometry:**
  • Write and balance the chemical equation.
  • Convert known substance amounts to moles.
  • Use mole ratios (from the equation) to calculate the number of moles of the desired substance.
  • Convert from moles to the required unit of measure (grams, liters, molecules, etc.).
Stoichiometry's precision allows chemists to create desired products without excess waste, which is both economically and environmentally beneficial.
Ion Counting
Ion counting is an integral part of chemistry that involves determining the quantity of ions present in a chemical compound. This is especially important in the context of compounds forming ionic bonds, where ions are present in a specific ratio.
For instance, in the exercise problem given, counting ions is crucial to determine how many ions are in a given number of moles of a substance.
**Examples:**
  • In the case of \( \mathrm{Na}_2\mathrm{S} \), there is one \( \mathrm{S}^{2-} \) ion per molecule, meaning the moles of \( \mathrm{S}^{2-} \) ions directly equal the moles of \( \mathrm{Na}_2\mathrm{S} \).
  • For \( \mathrm{Mg}_3(\mathrm{PO}_4)_2 \), there are three \( \mathrm{Mg}^{2+} \) ions for each molecule, so the number of moles must be multiplied by three to get the total moles of \( \mathrm{Mg}^{2+} \) ions.
  • In \( \mathrm{Ca}(\mathrm{CH}_3\mathrm{COO})_2 \), there are two acetate ions per molecule, requiring multiplication by two to find the total moles of \( \mathrm{CH}_3\mathrm{COO}^- \) ions.
Ion counting is crucial for understanding the properties of ionic compounds and how they will behave in chemical reactions.
Chemical Formulas
Chemical formulas are symbolic representations of chemical substances. They display the types of atoms and their ratios in a molecule, acting as a map for molecular composition. Understanding chemical formulas is key to solving problems involving mole calculations and ion counting.
**Key Elements of Chemical Formulas:**
  • Subscripts: These indicate the number of atoms for each element in a molecule (e.g., \( \mathrm{H}_2O \) shows two hydrogen atoms and one oxygen atom).
  • Parentheses: Used in complex molecules to denote groups of atoms that repeat (e.g., \( \mathrm{Mg}_3(\mathrm{PO}_4)_2 \) indicates two phosphate groups attached to three magnesium atoms).
  • Charges: Important for ionic compounds, showing overall charge balance (e.g., \( \mathrm{SO}_4^{2-} \)).
Understanding chemical formulas ensures that proper stoichiometry is applied during chemical calculations. This enables accurate predictions of reactant needs and product yields.

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Most popular questions from this chapter

Answer true or false. (a) Stoichiometry is the study of mass relationships in chemical reactions. (b) To determine mass relationships in a chemical reaction, you first need to know the balanced chemical equation for the reaction. (c) To convert from grams to moles and vice versa, use Avogadro's number as a conversion factor. (d) To convert from grams to moles and vice versa, use molar mass as a conversion factor. (e) A limiting reagent is the reagent that is used up first. (f) Suppose a chemical reaction between A and B requires 1 mol of \(A\) and 2 mol of \(B\). If 1 mol of each is present, then \(\mathrm{B}\) is the limiting reagent. (g) Theoretical yield is the yield of product that should be obtained according to the balanced chemical equation. (h) Theoretical yield is the yield of product that should be obtained if all limiting reagent is converted to product. (i) Percent yield is the number of grams of product divided by the number of grams of the limiting reagent times 100 (j) To calculate percent yield, divide the mass of product formed by the theoretical yield and multiply by 100

When a piece of aluminum metal is dropped into hydrochloric acid, HCl, hydrogen is released as a gas and a solution of aluminum chloride forms. Write a balanced equation for the reaction.

A sample of gold consisting of \(8.68 \times 10^{23}\) atoms with a density of \(19.3 \mathrm{g} / \mathrm{mL}\) is hammered into a sheet that covers an area of \(1.00 \times 10^{2} \mathrm{ft}^{2}\). Determine the thick- ness of the sheet in centimeters.

At one time, acetaldehyde was prepared industrially by the reaction of ethylene with air in the presence of a copper catalyst. How many grams of acetaldehyde can be prepared from 81.7 g of ethylene? \(2 \mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \stackrel{\mathrm{Catal}_{5} \mathrm{s}}{\longrightarrow} 2 \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}(\mathrm{g})\)

Lead(II) nitrate and aluminum chloride react according to the following equation: \(3 \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}+2 \mathrm{AlCl}_{3} \longrightarrow 3 \mathrm{PbCl}_{2}+2 \mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}\) In an experiment, 8.00 g of lead nitrate reacted with \(2.67 \mathrm{g}\) of aluminum chloride to give \(5.55 \mathrm{g}\) of lead chloride. (a) Which reactant was the limiting reagent? (b) What was the percent yield?

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