Problem 83 - General Chemistry: Principles and Modern Applications

Question

How ma475.15 grams \(\mathrm{HCl}\)ny grams of HCl are consumed in the reaction of \(425 \mathrm{g}\) of a mixture containing \(35.2 \% \mathrm{MgCO}_{3}\) and \(64.8 \% \mathrm{Mg}(\mathrm{OH})_{2},\) by mass? $$\begin{array}{c} \mathrm{Mg}(\mathrm{OH})_{2}+2 \mathrm{HCl} \longrightarrow \mathrm{MgCl}_{2}+2 \mathrm{H}_{2} \mathrm{O} \\ \mathrm{MgCO}_{3}+2 \mathrm{HCl} \longrightarrow \mathrm{MgCl}_{2}+\mathrm{H}_{2} \mathrm{O}+\mathrm{CO}_{2}(\mathrm{g}) \end{array}$$

Step-by-Step Solution

Answer

The total mass of HCl consumed in the reaction is 475.15 grams.

Step 1: Calculate the amount of each compound in the mixture

First, calculate the amount of each compound in the mixture by applying the percentage by mass. For \(MgCO_3\), we have \(425g * 0.352 = 149.6g\). For \(Mg(OH)_2\), we get \(425g * 0.648 = 275.4g\).

Step 2: Determine the number of moles of each compound

Next, convert the mass of each compound to the number of moles using the molar mass. The molar mass of \(MgCO_3\) is \(24.3g/mol + 12.01g/mol + 3*16.00g/mol = 84.31g/mol\) and that of \(Mg(OH)_2\) is \(24.3g/mol + 2*(1.01g/mol + 16.00g/mol) = 58.32g/mol\). Therefore, there are \(149.6g / 84.31g/mol = 1.774mol\) of \(MgCO_3\) and \(275.4g / 58.32g/mol = 4.723mol\) of \(Mg(OH)_2\).

Step 3: Apply the stoichiometric relationships

For each mole of \(MgCO_3\) and \(Mg(OH)_2\), two moles of HCl are consumed according to the balanced equation. This implies that \(1.774mol * 2 = 3.548mol\) of HCl react with \(MgCO_3\) and \(4.723mol * 2 = 9.446mol\) of HCl react with \(Mg(OH)_2\).

Step 4: Calculate the total mass of HCl

Finally, convert moles of HCl consumed to mass using molar mass of HCl (\(1.01g/mol + 35.45g/mol = 36.46g/mol\)). Hence, the total mass of HCl consumed is \((3.548mol + 9.446mol) * 36.46g/mol = 475.15g\).

Key Concepts

Chemical Reactions

Chemical reactions involve transforming reactants into products, following specific patterns and rules. In the given problem, two reactions are of interest:

• Reaction 1: \( \mathrm{Mg(OH)}_2 + 2 \mathrm{HCl} \rightarrow \mathrm{MgCl}_2 + 2 \mathrm{H}_2\mathrm{O} \)
• Reaction 2: \( \mathrm{MgCO}_3 + 2 \mathrm{HCl} \rightarrow \mathrm{MgCl}_2 + \mathrm{H}_2\mathrm{O} + \mathrm{CO}_2 \)

Both reactions consume hydrochloric acid (HCl) and produce different products. The equations must be balanced to reflect the conservation of mass, meaning the number of each atom on the reactant side equals that on the product side. In these reactions, two moles of HCl are consumed per mole of either \( \mathrm{Mg(OH)}_2 \) or \( \mathrm{MgCO}_3 \).
Understanding these reactions helps predict the quantities of products formed. Knowing chemical reactions is fundamental in stoichiometry, allowing us to calculate reactants and products needed or produced under controlled conditions.

Molar Mass Calculation

Molar mass is vital for converting between grams and moles. It refers to the mass of one mole of a substance, typically measured in grams per mole (g/mol). Calculations often involve summing the atomic masses of all atoms in a single molecule of the compound.
For instance, the molar mass of \( \mathrm{MgCO}_3 \) is derived from its atomic components:

• Magnesium (Mg): 24.3 g/mol
• Carbon (C): 12.01 g/mol
• Oxygen (O): 3 x 16.00 g/mol

This adds up to a total molar mass of 84.31 g/mol. Similarly, \( \mathrm{Mg(OH)}_2 \) has a molar mass calculated as:

• Magnesium (Mg): 24.3 g/mol
• Hydrogen (H): 2 x 1.01 g/mol
• Oxygen (O): 2 x 16.00 g/mol

Making its total molar mass 58.32 g/mol.
These calculations are essential for converting the mass of a substance to moles, thereby facilitating stoichiometric calculations in chemical equations.

Mass Percent

Mass percent indicates the concentration of a component in a mixture, expressed as the mass of a specific component divided by the total mass of the mixture, multiplied by 100. It is a simple and effective way to describe the composition of mixtures and solutions.
In this exercise, the mixture comprised 35.2% \( \mathrm{MgCO}_3 \) and 64.8% \( \mathrm{Mg(OH)}_2 \). To find the grams of each compound in a 425 g sample:

• Mass of \( \mathrm{MgCO}_3 \): 425 g x 0.352 = 149.6 g
• Mass of \( \mathrm{Mg(OH)}_2 \): 425 g x 0.648 = 275.4 g

Mass percent is crucial in chemical calculations, especially when determining the amounts of substances needed for reactions or analyzing product yields.