Chapter 4: Problem 57
What mass of solid aluminum hydroxide can be produced when \(50.0 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}\) is added to \(200.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{KOH} ?\)
Short Answer
Expert verified
The mass of solid aluminum hydroxide that can be produced when \(50.0 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{Al(NO_3)_3}\) is added to \(200.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{KOH}\) is \(0.780 \mathrm{~g}\).
Step by step solution
01
Write the balanced chemical equation
Firstly, we need to write the balanced chemical equation for the reaction between aluminum nitrate and potassium hydroxide.
\[ \mathrm{Al(NO_3)_3 + 3KOH \rightarrow Al(OH)_3 + 3KNO_3} \]
02
Determine the moles of reactants
Next, we need to determine the moles of both reactants using their volumes and molarities. Given the volumes and molarities of the solutions, we can calculate the moles by multiplying the volume (in liters) and the molarity of the solution.
Moles of \(\mathrm{Al(NO_3)_3} = 50.0 \mathrm{~mL} \times 0.200 \mathrm{M} = 0.050 \mathrm{~L} \times 0.200 \mathrm{M} = 0.0100 \mathrm{~mol}\)
Moles of \(\mathrm{KOH} = 200.0 \mathrm{~mL} \times 0.100 \mathrm{M} = 0.200 \mathrm{~L} \times 0.100 \mathrm{M} = 0.0200 \mathrm{~mol}\)
03
Determine the limiting reactant
To find out which reactant is limiting, we have to compare the mole ratios of each reactant to their stoichiometric coefficients in the balanced chemical equation. As per the balanced equation, 1 mol of \(\mathrm{Al(NO_3)_3}\) reacts with 3 mol \(\mathrm{KOH}\). Using mole ratio, we can determine how much of each reactant would be consumed.
Moles of \(\mathrm{KOH}\) required to react with the given moles of \(\mathrm{Al(NO_3)_3} = \frac{1}{3} \times 0.0100 \mathrm{~mol} = 0.00333 \mathrm{~mol}\)
Since there are more moles of \(\mathrm{KOH}\) (0.0200 mol) in the solution than required (0.00333 mol), this means that \(\mathrm{Al(NO_3)_3}\) is the limiting reactant.
04
Calculate the mass of aluminum hydroxide
As per the balanced equation, 1 mol of \(\mathrm{Al(NO_3)_3}\) produces 1 mol of aluminum hydroxide (\(\mathrm{Al(OH)_3}\)). Since we have 0.0100 mol of the limiting reactant, this will also be the moles of aluminum hydroxide produced.
Moles of \(\mathrm{Al(OH)_3} = 0.0100 \mathrm{~mol}\)
Now, we need to convert moles to mass using the molar mass of \(\mathrm{Al(OH)_3\). The molar mass of \(\mathrm{Al(OH)_3 = 27 + (3 \times 16) + (3 \times 1) = 27 + 48 + 3 = 78 \mathrm{~g/mol}\). To find the mass of aluminum hydroxide, we do the following:
Mass of \(\mathrm{Al(OH)_3 = 0.0100 \mathrm{~mol} \times 78 \mathrm{~g/mol} = 0.780 \mathrm{~g}\)
05
Present the answer
The mass of solid aluminum hydroxide that can be produced when \(50.0 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{Al(NO_3)_3}\) is added to \(200.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{KOH}\) is \(0.780 \mathrm{~g}\).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with Vaia!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Limiting Reactant
In any chemical reaction, the limiting reactant is the substance that is completely consumed first, limiting the amount of product that can be formed. It's the reactant that determines the maximum yield of products.
To identify the limiting reactant, you must compare the amount of reactants used in the reaction with the stoichiometry of the chemical equation. In practice, this compares the mole ratios of the reactants with the coefficients of the balanced equation. If you have more of a reactant than is required by these ratios, it is in excess; the one you have less of in comparison to its required stoichiometric amount is the limiting reactant. Knowing this concept is crucial because it allows you to calculate the correct amount of products that can form.
To identify the limiting reactant, you must compare the amount of reactants used in the reaction with the stoichiometry of the chemical equation. In practice, this compares the mole ratios of the reactants with the coefficients of the balanced equation. If you have more of a reactant than is required by these ratios, it is in excess; the one you have less of in comparison to its required stoichiometric amount is the limiting reactant. Knowing this concept is crucial because it allows you to calculate the correct amount of products that can form.
Mole Ratio
The mole ratio is another fundamental concept of stoichiometry and it comes from the balanced chemical equation. It refers to the ratio of moles of one substance to the moles of another substance in a balanced equation. These ratios allow chemists to calculate the amounts of reactants needed and the amount of product that can be formed during a reaction.
For example, if the balanced chemical equation indicates that 2 moles of hydrogen gas react with 1 mole of oxygen gas to produce 2 moles of water, then the mole ratio of hydrogen to oxygen is 2:1. This quantifies the precise proportions in which chemicals react, and it helps in determining which reactant is the limiting reactant.
For example, if the balanced chemical equation indicates that 2 moles of hydrogen gas react with 1 mole of oxygen gas to produce 2 moles of water, then the mole ratio of hydrogen to oxygen is 2:1. This quantifies the precise proportions in which chemicals react, and it helps in determining which reactant is the limiting reactant.
Balanced Chemical Equation
A balanced chemical equation is another cornerstone of stoichiometry. It represents a chemical reaction where the number of atoms for each element is equal on the reactant and the product sides. This principle, known as the law of conservation of mass, ensures that matter is neither created nor destroyed during a chemical reaction.
Each component is represented by its chemical formula, and coefficients are used to show the number of molecules or moles of each substance involved. It is imperative to balance a chemical equation to accurately perform calculations related to the reaction, such as determining the limiting reactant or the mole ratios needed for complete reaction.
Each component is represented by its chemical formula, and coefficients are used to show the number of molecules or moles of each substance involved. It is imperative to balance a chemical equation to accurately perform calculations related to the reaction, such as determining the limiting reactant or the mole ratios needed for complete reaction.
Molar Mass
Finally, molar mass is a concept that relates the mass of a substance to the number of moles present. It is defined as the mass of a given substance (chemical element or chemical compound) divided by its amount of substance in moles. The molar mass is typically expressed in grams per mole (g/mol) and is calculated by adding the atomic masses of all atoms present in the molecule.
For stoichiometric calculations, the molar mass allows conversion between moles and grams. In the example provided, we use the molar mass of aluminum hydroxide to convert the moles of product predicted by the stoichiometry into the mass of product that can actually be formed from the limiting reactant.
For stoichiometric calculations, the molar mass allows conversion between moles and grams. In the example provided, we use the molar mass of aluminum hydroxide to convert the moles of product predicted by the stoichiometry into the mass of product that can actually be formed from the limiting reactant.