Question

For each of the following unbalanced chemical equations, suppose \(1.00 \mathrm{g}\) of each reactant is taken. Show by calculation which reactant is limiting. Calculate the mass of each product that is expected. a. \(\mathrm{UO}_{2}(s)+\mathrm{HF}(a q) \rightarrow \mathrm{UF}_{4}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) b. \(\mathrm{NaNO}_{3}(a q)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow\) \(\mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+\mathrm{HNO}_{3}(a q)\) c. \(\mathrm{Zn}(s)+\mathrm{HCl}(a q) \rightarrow \mathrm{ZnCl}_{2}(a q)+\mathrm{H}_{2}(g)\) d. \(\mathrm{B}(\mathrm{OH})_{3}(s)+\mathrm{CH}_{3} \mathrm{OH}(l) \rightarrow \mathrm{B}\left(\mathrm{OCH}_{3}\right)_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l)\)

Short answer

a. Limiting reactant: UO₂. Expected mass of products: 0.99 g UF₄ and 0.133 g H₂O. b. Limiting reactant: NaNO₃. Expected mass of products: 0.48 g Na₂SO₄ and 0.54 g HNO₃. c. Limiting reactant: Zn. Expected mass of products: 0.423 g ZnCl₂ and 0.0123 g H₂. d. Limiting reactant: B(OH)₃. Expected mass of products: 0.410 g B(OCH₃)₃ and 0.550 g H₂O.

Step-by-step solution

Calculate moles of each reactant

Calculate the moles of each reactant using their molar mass: UO₂: \( \frac{1.00 \ \text{g}}{270.03 \ \text{g/mol}} = 0.00370 \ \text{mol} \) HF: \( \frac{1.00 \ \text{g}}{20.01 \ \text{g/mol}} = 0.04998 \ \text{mol} \)

Identify the limiting reactant

Compare the mole ratio of UO₂ to HF (1:4). Then determine which is the limiting reactant by dividing the moles of each reactant by their stoichiometric coefficient: UO₂: \( \frac{0.00370 \ \text{mol}}{1} = 0.00370 \) HF: \( \frac{0.04998 \ \text{mol}}{4} = 0.01249 \) As UO₂ has a smaller value, it is the limiting reactant.

Calculate mass of each product

Using the limiting reactant UO₂, calculate the mass of each product: UF₄: \(0.00370 \ \text{mol} \times \frac{1 \ \text{mol} \ \mathrm{UF_{4}}}{1 \ \text{mol} \ \mathrm{UO_{2}}} \times 268.04 \ \text{g/mol} = 0.99 \ \text{g}\) H₂O: \(0.00370 \ \text{mol} \times \frac{2 \ \text{mol} \ \mathrm{H_{2}O}}{1 \ \text{mol} \ \mathrm{UO_{2}}} \times 18.015 \ \text{g/mol} = 0.133 \ \text{g}\) We can expect 0.99 g of UF₄ and 0.133 g of H₂O. Repeat these steps for the other reactions: b. \(2\mathrm{NaNO}_{3}(a q)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow\) \(\mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+2\mathrm{HNO}_{3}(a q)\) NaNO₃: 0.01261 mol H₂SO₄: 0.01033 mol Limiting reactant: NaNO₃ Expected mass of products: 0.48 g Na₂SO₄, 0.54 g HNO₃ c. \(\mathrm{Zn}(s)+2\mathrm{HCl}(a q) \rightarrow \mathrm{ZnCl}_{2}(a q)+\mathrm{H}_{2}(g)\) Zn: 0.01534 mol HCl: 0.02742 mol Limiting reactant: Zn Expected mass of products: 0.423 g ZnCl₂, 0.0123 g H₂ d. \(\mathrm{B}(\mathrm{OH})_{3}(s)+3\mathrm{CH}_{3} \mathrm{OH}(l) \rightarrow \mathrm{B}\left(\mathrm{OCH}_{3}\right)_{3}(s)+3\mathrm{H}_{2} \mathrm{O}(l)\) B(OH)₃: 0.01020 mol CH₃OH: 0.03115 mol Limiting reactant: B(OH)₃ Expected mass of products: 0.410 g B(OCH₃)₃, 0.550 g H₂O

Key concepts

Stoichiometry: The Heart of Chemical Calculations

Understanding stoichiometry is crucial when dealing with chemical reactions. It's like the recipe in a kitchen that tells you how much of each ingredient you need. Chemical reactions have reactants and products, and stoichiometry helps us identify the correct proportions and relationships between them.

In essence, stoichiometry allows us to use the coefficients in balanced chemical equations to determine how much of each substance participates in the reaction. This is important for finding the limiting reactant, which is the reactant that is completely consumed first, limiting the amount of products formed.

Step-by-step, stoichiometry involves:

• Calculating the moles of each reactant present in the mixture.
• Using the coefficients in the balanced equation to see how many moles of each reactant are needed.
• Identifying which reactant runs out first, as this one limits the reaction.

With these steps, the stoichiometric calculations allow us to predict the amount of product that can form, making it a key concept in any chemical reaction analysis.

Chemical Equations: Blueprint for Reactions

Chemical equations serve as a blueprint for chemical reactions. They are symbolic representations showing what happens to elements and compounds during a chemical reaction. They list reactants (beginning substances) and products (end substances), along with their quantities.

Understanding how to read and write chemical equations is vital because it ensures that all atoms are accounted for through the Law of Conservation of Mass. This law states that atoms cannot be created or destroyed in chemical reactions, so the same number of each type of atom must be present on both sides of the equation.

Balancing a chemical equation involves adjusting the coefficients (the numbers before molecules) to get an equal number of each type of atom on both sides. For instance, if a reaction shows 1 mole of \("UO_2"\) reacting with 4 moles of \("HF"\), then the equation is balanced when the products reflect these proportions.

Balanced equations are a crucial tool for stoichiometry and limiting reactant calculations, as they provide the necessary ratios of reactants to products.

Molar Mass Calculation: The Weight of Molecules

Calculating molar mass is like weighing the ingredients for your chemical recipe. Molar mass is the mass of one mole of a substance, expressed in grams per mole (g/mol). It's a fundamental concept for converting between grams and moles, which is essential in stoichiometry.

To calculate the molar mass, you need to sum up the atomic masses (from the periodic table) of all the atoms in a formula. For example, the molar mass of \("UO_2"\) is calculated by adding together the molar masses of uranium and oxygen: \(238.03 \text{ g/mol (U)} + 2 \times 16.00 \text{ g/mol (O)} = 270.03 \text{ g/mol}\).

This calculation is vital because:

• It allows conversion between the mass of a substance and the number of moles.
• It helps determine how much of a reactant is present in a given mass, which is crucial for finding the limiting reactant.
• It is necessary for calculating the expected yield of a product from known amounts of reactants.

By mastering molar mass calculations, you gain the ability to decode the quantitative aspects of chemical reactions.