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Chapter 1: Problem 173

\(320 \mathrm{mg}\) of a sample of magnesium having a coating of its oxide required \(20 \mathrm{~mL}\) of \(0.1 \mathrm{M}\) hydrochloric acid for the complete neutralisation of the latter. The composition of the sample is: (a) \(87.5 \% \mathrm{Mg}\) and \(12.5 \% \mathrm{MgO}\) (b) \(12.5 \% \mathrm{Mg}\) and \(87.5 \% \mathrm{MgO}\) (c) \(80 \% \mathrm{Mg}\) and \(20 \% \mathrm{MgO}\) (d) \(20 \% \mathrm{Mg}\) and \(80 \% \mathrm{MgO}\)

Short Answer

Expert verified
The composition of the sample is 87.5% Mg and 12.5% MgO.

Step by step solution

01

Calculate moles of HCl

Firstly, calculate the moles of HCl that reacted with the sample. Use the molarity (M) and volume (L) relationship: moles = molarity \(\times\) volume. Given that the volume of HCl is 20 mL (or 0.020 L) and the molarity is 0.1 M, the moles of HCl are: \(0.1 \text{ M} \times 0.020 \text{ L} = 0.002 \text{ moles} \).
02

Write the chemical reaction

Write the balanced chemical reaction between magnesium oxide (MgO) and hydrochloric acid (HCl). The balanced equation is: \( \text{MgO} + 2\text{HCl} \rightarrow \text{MgCl}_2 + \text{H}_2\text{O} \).
03

Calculate moles of MgO using the balanced equation

Using the stoichiometry of the balanced reaction, calculate the moles of MgO that reacted with HCl. The ratio of MgO to HCl is 1:2, so the moles of MgO is half the moles of HCl. Therefore, the moles of MgO is \(0.002 \text{ moles HCl} \div 2 = 0.001 \text{ moles MgO}\).
04

Calculate the mass of MgO

Use the molar mass of MgO (24.305 + 15.999 = 40.304 g/mol) to calculate the mass of the MgO that reacted: \(0.001 \text{ moles MgO} \times 40.304 \text{ g/mol} = 0.040304 \text{ g} = 40.304 \text{ mg}\).
05

Determine the percentage composition

Calculate the percentage of MgO in the sample by dividing the mass of MgO by the total mass of the sample and multiplying by 100%. The percentage of Mg is then the difference between 100% and the percentage of MgO. Given the total mass of the sample is 320 mg: \( \text{Percentage of MgO} = \frac{40.304 \text{ mg}}{320 \text{ mg}} \times 100\% = 12.595\% \). Approximate to 12.5% for MgO and 87.5% for Mg.

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Key Concepts

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

Understanding Chemical Reaction Equations
Chemical reaction equations are the language through which chemists communicate how substances react. These equations are balanced to obey the law of conservation of mass, meaning the number of atoms of each element on the reactant side must equal those on the product side.

For example, the reaction between magnesium oxide (MgO) and hydrochloric acid (HCl) can be written as: \[ \text{MgO} + 2\text{HCl} \rightarrow \text{MgCl}_2 + \text{H}_2\text{O} \]In this equation, for every one molecule of MgO, two molecules of HCl are needed to form one molecule of magnesium chloride (MgCl2) and one molecule of water (H2O). Understanding this stoichiometric ratio is pivotal for solving problems involving chemical reactions, as it dictates the proportions in which reactants combine to form products.
The Role of Molarity and Concentration in Stoichiometry
Molarity, a measure of concentration, expresses the number of moles of a solute present in one liter of solution. It's crucial for quantifying the substances in a chemical reaction. The formula to calculate the number of moles from molarity is: \[ \text{Moles} = \text{Molarity} \times \text{Volume} \]where the volume must be in liters.

In the given problem, the molarity of HCl is used alongside the volume of HCl solution to find the number of moles of HCl that participate in the reaction: \[ 0.1 \text{ M} \times 0.020 \text{ L} = 0.002 \text{ moles of HCl} \]Grasping how to convert between moles, molarity, and volume is essential for students studying chemistry, as it's the foundation for understanding how much of each reactant is used in a reaction.
Calculating Percentage Composition
Percentage composition calculation is a fundamental concept in stoichiometry, allowing chemists to quantify the relative amounts of elements or compounds in a mixture. The percentage of a component is calculated by dividing the mass of that component by the total mass of the mixture, then multiplying by 100%.

For instance, in the provided solution: \[ \text{Percentage of MgO} = \frac{40.304 \text{ mg}}{320 \text{ mg}} \times 100\% = 12.595\% \]Thus, after rounding, the sample consists of approximately 12.5% magnesium oxide and the rest magnesium. Understanding how to calculate percentage composition underpins many applications, such as determining the purity of a sample or the formulation of mixtures in industry.

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Most popular questions from this chapter

\(1 \mathrm{~g}\) mixture of equal number of mole of \(\mathrm{Li}_{2} \mathrm{CO}_{3}\) and other metal carbonate \(\left(\mathrm{M}_{2} \mathrm{CO}_{3}\right)\) 1e \(21.6 \mathrm{~mL}\) of \(0.5 \mathrm{~N} \mathrm{HCl}\) for complete neutralisation reaction. What is the approximate atomic weight of the other metal ? (a) 25 (b) 23 (c) 24 (d) 51

A sample of ammonium phosphate, \(\left(\mathrm{NH}_{4}\right)_{3} \mathrm{PO}_{4}\), contains 6 moles of hydrogen atoms. The number of moles of oxygen atoms in the sample is: (a) 1 (b) 2 (c) 4 (d) 6

\(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) solution of 1 molal concentration is present in 1 litre solution of \(2.684 \mathrm{~g} / \mathrm{cc} .\) How many moles of \(\mathrm{BaSO}_{4}\) would be precipitated on adding \(\mathrm{BaCl}_{2}\) in excess ? (a) 2 moles (b) 3 moles (c) 6 moles (d) 12 moles

If \(m_{A}\) gram of a metal \(A\) displaces \(m_{B}\) gram of another metal \(B\) from its salt solution and if the equivalent weights are \(E_{A}\) and \(E_{B}\) respectively then equivalent weight of \(A\) can be expressed \(\begin{aligned}&\text { as: } \\\&\text { (a) } E_{A}=\frac{m_{A}}{m_{B}} \times E_{B} \quad \text { (b) } E_{A}=\frac{m_{A} \times m_{B}}{E_{B}} \quad \text { (c) } E_{A}=\frac{m_{B}}{m_{A}} \times E_{B} & \text { (d) } E_{A}=\sqrt{\frac{m_{A}}{m_{B}} \times E_{B}}\end{aligned}\)

A mixture of \(\mathrm{O}_{2}\) and gas " \(Y\) " (mol. wt. 80) in the mole ratio \(a: b\) has a mean molecular weight 40\. What would be mean molecular weight, if the gases are mixed in the ratio \(b: a\) under identical conditions ? (gases are non-reacting): (a) 40 (b) 48 (c) 62 (d) 72
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