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Chapter 15: Problem 34

Determine how many grams of \(\mathrm{Al}(\mathrm{OH})_{3}\) will be required to neutralize \(275 \mathrm{~mL}\) of \(0.125 \mathrm{M} \mathrm{HCl}\) according to the reaction: $$ 3 \mathrm{HCl}(a q)+\mathrm{Al}(\mathrm{OH})_{3}(s) \longrightarrow \mathrm{AlCl}_{3}(a q)+3 \mathrm{H}_{2} \mathrm{O}(l) $$

Short Answer

Expert verified
0.894 g of Al(OH)3 are required.

Step by step solution

01

- Identify the given information

Volume of \(\text{HCl}\) solution = 275 mL = 0.275 L \(\text{Molarity of HCl solution}\) = 0.125 M
02

- Calculate moles of HCl

Use the formula \( \text{Moles of HCl} = \text{Molarity of HCl} \times \text{Volume of HCl in Liters} \). This yields \[ 0.125 \times 0.275 = 0.034375 \text{ moles of HCl} \]
03

- Use stoichiometry to find moles of Al(OH)3 needed

From the balanced equation, 3 moles of HCl react with 1 mole of Al(OH)3. Therefore, the moles of Al(OH)3 needed are \[ \text{Moles of Al(OH)3} = \frac{\text{Moles of HCl}}{3} = \frac{0.034375}{3} = 0.01145833 \text{ moles} \]
04

- Calculate the mass of Al(OH)3

Find the molar mass of Al(OH)3: \(\text{Molar mass} = (1 \times \text{Al}) + (3 \times \text{O}) + (3 \times \text{H})\) = 26.98 + (3 \times 16.00) + (3 \times 1.01) = 78.00 \text{ g/mol}. \ \text{Mass of Al(OH)3}= \text{Moles of Al(OH)3} \times \text{Molar mass of Al(OH)3} \ \text{Mass} = 0.01145833 \times 78.00 = 0.89375 \text{ g}.

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Key Concepts

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

stoichiometry
Stoichiometry is the calculation of reactants and products in chemical reactions. It helps you understand the proportions in which chemicals react. In our given problem, the balanced chemical equation shows the relationship between \(\text{HCl}\) and \(\text{Al(OH)}_{3}\). Specifically, 3 moles of \(\text{HCl}\) react with 1 mole of \(\text{Al(OH)}_{3}\). This information lets us use the mole ratio from our balanced equation to find how many moles of \(\text{Al(OH)}_{3}\) are needed to neutralize the given moles of \(\text{HCl}\). Ensuring you understand stoichiometry is essential for any chemical calculation!
molarity
Molarity, often represented as \(\text{M}\), describes the concentration of a solution. It is defined as the number of moles of solute per liter of solution: \[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}} \] In our problem, we have a 0.125 M solution of HCl, which means there are 0.125 moles of HCl in every liter of solution. Knowing the molarity and volume—in this case 0.275 L—allows us to determine how many moles of HCl are available for the reaction, which is crucial for the next steps in stoichiometry.
moles calculation
Calculating moles is fundamental in chemistry. It connects the macroscopic world we can see with the microscopic world of atoms and molecules. The formula used is: \[ \text{Moles} = \text{Molarity} \times \text{Volume (in Liters)} \] From our exercise, using the molarity of the HCl solution (0.125 M) and the volume (0.275 L), we calculated: \[ \text{Moles of HCl} = 0.125 \times 0.275 = 0.034375 \text{ moles} \] These moles are then used to determine how many moles of \(\text{Al(OH)}_{3}\) will react by using the stoichiometric ratios from the balanced chemical equation.
balanced chemical equation
A balanced chemical equation shows the proportion of reactants and products in a reaction. This is vital to ensure the law of conservation of mass is followed. For our example reaction: \(\text{3 HCl (aq) + Al(OH)}_{3} \text{(s)} \rightarrow \text{AlCl}_{3} \text{(aq) + 3 H}_{2}\text{O} \text{(l)}\), it is essential to have each side balanced with the same number of atoms for each element. This tells us that three moles of HCl react with one mole of Al(OH)3. Thus, by knowing the amount of HCl, we can determine the required amount of Al(OH)3 using simple stoichiometric ratios.

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Most popular questions from this chapter

Determine whether each of the following substances is an electrolyte or a nonelectrolyte. All are mixed with water. (a) \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) (glucose) (d) \(\mathrm{LiOH}\) (b) \(\mathrm{P}_{2} \mathrm{O}_{5}\) (e) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) (ethyl alcohol) (c) \(\mathrm{NaClO}\) (f) \(\mathrm{KMnO}_{4}\)

Determine the molarity of each of the ions present in the following aqueous salt solutions: (assume \(100 \%\) ionization) (a) \(1.25 M \mathrm{CuBr}_{2}\) (b) \(0.75 \mathrm{M} \mathrm{NaHCO}_{3}\) (c) \(3.50 M \mathrm{~K}_{3} \mathrm{AsO}_{4}\) (d) \(0.65 M\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\)

A \(10.0-\mathrm{mL}\) sample of \(\mathrm{HNO}_{3}\) was diluted to a volume of \(100.00 \mathrm{~mL}\). Then \(25 \mathrm{~mL}\) of that diluted solution were needed to neutralize \(50.0 \mathrm{~mL}\) of \(0.60 \mathrm{M} \mathrm{KOH}\). What was the concentration of the original nitric acid?

Balance each of the following equations; then change them into balanced net ionic equations: (a) \(\mathrm{Mg}(s)+\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(a q) \longrightarrow \mathrm{Cu}(s)+\mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}(a q)\) (b) \(\mathrm{H}_{2} \mathrm{SO}_{4}(a q)+\mathrm{NaOH}(a q) \longrightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) (c) \(\mathrm{K}_{2} \mathrm{SO}_{3}(a q)+\mathrm{HClO}_{4}(a q) \longrightarrow\) \(\mathrm{KClO}_{4}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{SO}_{2}(g)\)

Write an equation showing how sodium sulfate dissociates in water.
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