Question

A mixture of bases can sometimes be the active ingredient in antacid tablets. If \(0.4826 \mathrm{~g}\) of a mixture of \(\mathrm{Al}(\mathrm{OH})_{3}\) and \(\mathrm{Mg}(\mathrm{OH})_{2}\) is neutralized with \(17.30 \mathrm{~mL}\) of \(1.000 \mathrm{M} \mathrm{HNO}_{3}\), what is the mass \(\%\) of \(\mathrm{Al}(\mathrm{OH})_{3}\) in the mixture?

Short answer

Mass \(\%\) of \(\mathrm{Al}(\mathrm{OH})_3\) is calculated using simultaneous equations to find moles, then converting to mass percentage.

Step-by-step solution

Write the balanced chemical equations

First, write the balanced chemical equations for the neutralization reactions involving \(\mathrm{Al}(\mathrm{OH})_3\) and \(\mathrm{Mg}(\mathrm{OH})_2\): \[\mathrm{Al}(\mathrm{OH})_3 + 3\mathrm{HNO}_3 \rightarrow \mathrm{Al}(\mathrm{NO}_3)_3 + 3\mathrm{H}_2\mathrm{O}\ \mathrm{Mg}(\mathrm{OH})_2 + 2\mathrm{HNO}_3 \rightarrow \mathrm{Mg}(\mathrm{NO}_3)_2 + 2\mathrm{H}_2\mathrm{O}\].

Calculate the moles of HNO₃

Using the given volume and molarity of the \(\mathrm{HNO}_3\) solution, calculate the number of moles: \(\text{moles of } \mathrm{HNO}_3 = 1.000 \, \mathrm{M} \times 17.30 \, \mathrm{mL} = 0.01730 \, \mathrm{mol}\).

Set up the equations for moles used

Let \(x\) be the moles of \(\mathrm{Al}(\mathrm{OH})_3\) and \(y\) be the moles of \(\mathrm{Mg}(\mathrm{OH})_2\). The total moles of \(\mathrm{HNO}_3\) are used up by both reactions: \(3x + 2y = 0.01730\).

Relate moles to mass

Write the expressions for the total mass of the mixture. Using molar masses: \[x \times 78.00 + y \times 58.32 = 0.4826 \, \mathrm{g}\], where 78.00 \, \mathrm{g/mol} is the molar mass of \(\mathrm{Al}(\mathrm{OH})_3\) and 58.32 \, \mathrm{g/mol} is the molar mass of \(\mathrm{Mg}(\mathrm{OH})_2\).

Solve the simultaneous equations

Using the equations from Steps 3 and 4, solve for \(x\) and \(y\): \[3x + 2y = 0.01730\, \frac{x}{78.00} + \frac{y}{58.32} = 0.4826\]. Calculate and find the values of \(x\) and \(y\).

Calculate the mass \(\%\) of \(\mathrm{Al}(\mathrm{OH})_3\)

Once \(x\) (moles of \(\mathrm{Al}(\mathrm{OH})_3\)) is found, convert it to mass: \(\text{mass of } \mathrm{Al}(\mathrm{OH})_3 = x \times 78.00\). Then, compute the mass percentage: \[\% \mathrm{Al}(\mathrm{OH})_3 = \frac{\text{mass of } \mathrm{Al}(\mathrm{OH})_3}{0.4826} \times 100\].

Key concepts

molar mass

At the core of many chemical calculations is the concept of molar mass. The molar mass of a compound is the mass of one mole of its molecules or ions.
It’s measured in grams per mole (g/mol). To find it, you sum the atomic masses of all the atoms in a molecule. For example, for aluminum hydroxide, \(\text{Al(OH)}_3\), you add the masses of 1 aluminum atom (Al), 3 oxygen atoms (O), and 3 hydrogen atoms (H).
Hence: \(\text{Molar mass of Al(OH)}_3 = 26.98 + 3*(15.999) + 3*(1.008) = 78.00 \text{g/mol}\).
Knowing the molar mass allows us to convert between grams and moles, a critical step for many calculations.

stoichiometry

Stoichiometry involves the quantitative relationships between reactants and products in a chemical reaction.
It allows us to use balanced chemical equations to predict the amounts of substances consumed and produced.
In our given exercise, the stoichiometry of the neutralization reactions tells us the ratio in which reactants combine with acids. For instance, \(\text{Al(OH)}_3\) reacts with \(\text{HNO}_3\) in a 1:3 ratio: \(\text{Al(OH)}_3 + 3\text{HNO}_3 \rightarrow \text{Al(NO}_3\text{)}_3 + 3\text{H}_2\text{O}\).
Therefore, 1 mole of \(\text{Al(OH)}_3\) will react with 3 moles of \(\text{HNO}_3\). These ratios help determine how much acid is needed to neutralize a given amount of base.

acid-base reaction

An acid-base reaction is a neutralization process where an acid and a base react to form water and a salt.
In the exercise, \(\text{HNO}_3\) (nitric acid) is reacting with the bases \(\text{Al(OH)}_3\) (aluminum hydroxide) and \(\text{Mg(OH)}_2\) (magnesium hydroxide).
These reactions can be written as: \(\text{Al(OH)}_3 + 3\text{HNO}_3 \rightarrow \text{Al(NO}_3\text{)}_3 + 3\text{H}_2\text{O}\) and \(\text{Mg(OH)}_2 + 2\text{HNO}_3 \rightarrow \text{Mg(NO}_3\text{)}_2 + 2\text{H}_2\text{O}\).
By breaking down the reactions, we see that water (H2O) and aluminum nitrate or magnesium nitrate are formed. It's essential to recognize the 1:3 and 1:2 molar ratios for neutralization in these reactions.

simultaneous equations

To solve problems involving mixtures like in our exercise, we often need to solve simultaneous equations. This involves finding the values of multiple variables that satisfy more than one equation at the same time.
In our specific problem, we use the stoichiometric relationships and masses to set up these equations.
For example, we know: \(\text{3x + 2y = 0.01730}\) (from the total moles of \(\text{HNO}_3\))
and \(\text{78.00x + 58.32y = 0.4826}\) (from the total mass of the mixture).
By solving these simultaneous equations, we find the molar quantities of \(\text{Al(OH)}_3\) (x) and \(\text{Mg(OH)}_2\) (y) to eventually determine their mass percentages in the mixture.