Question

For cach of the following unbalanced chemical cquations, 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. \(B(O H)_{3}(s)+C H_{3} O H(l) \rightarrow B\left(O C H_{3}\right)_{3}(s)+H_{2} O(l)\)

Short answer

For each of the given reactions, we first balance the chemical equation, convert the mass of reactants to moles, and then use stoichiometry to identify the limiting reactant and calculate the mass of the products expected. a. The balanced chemical equation is \(UO_{2}(s)+4 HF(aq) \rightarrow UF_{4}(aq)+2 H_{2}O(l)\). We find the limiting reactant to be \(UO_2\) with expected product masses of 0.888 g UF4 and 0.130 g H2O. b. The balanced chemical equation is \(2 NaNO_{3}(aq) + H_{2}SO_{4}(aq) \rightarrow Na_{2}SO_{4}(aq) + 2 HNO_{3}(aq)\). Follow a similar process to find the limiting reactant and the mass of the products expected. c. The balanced chemical equation is \(Zn(s) + 2 HCl(aq) \rightarrow ZnCl_{2}(aq) + H_{2}(g)\). Repeat the process to find the limiting reactant and the mass of the products expected. d. The balanced chemical equation is \(B(OH)_{3}(s) + 3 CH_{3}OH(l) \rightarrow B(OC H_{3})_{3}(s) + 3 H_{2}O(l)\). Carry out the same process to find the limiting reactant and the mass of the products expected.

Step-by-step solution

Balance the chemical equation

The balanced chemical equation is: \(UO_{2}(s)+4 HF(aq) \rightarrow UF_{4}(aq)+2 H_{2}O(l)\)

Convert mass of each reactant to moles

Using the molar masses: 1.00 g of \(UO_2 = \frac{1.00}{238 + 2\times16} \) mol ≈ 0.00361 mol 1.00 g of \(HF = \frac{1.00}{1 + 19} \) mol ≈ 0.0526 mol

Determine stoichiometry

UO2 : HF = 1 : 4

Calculate the amount of product formed from each reactant

From UO2: 0.00361 mol UO2 → 0.00361 mol UF4 (1 : 1 stoichiometry) From HF: 0.0526 mol HF → 0.0132 mol UF4 (4 : 1 stoichiometry)

Identify the limiting reactant

Since using UO2 gives a lower amount of UF4, UO2 is the limiting reactant.

Calculate the mass of products expected (UF4 and H2O)

From UO2: 0.00361 mol UF4 = 0.00361 × (238 + 4×19) g ≈ 0.888 g of UF4 expected 0.00361 mol UO2 yields 0.00722 mol H2O (2×0.00361), so: 0.00722 mol H2O = 0.00722 × (2 + 16) g ≈ 0.130 g of H2O expected The expected mass of products are: 0.888 g UF4 and 0.130 g H2O. Repeat the same process for the other reactions. b. \(2 NaNO_{3}(aq) + H_{2}SO_{4}(aq) \rightarrow Na_{2}SO_{4}(aq) + 2 HNO_{3}(aq)\) c. \(Zn(s) + 2 HCl(aq) \rightarrow ZnCl_{2}(aq) + H_{2}(g)\) d. \(B(OH)_{3}(s) + 3 CH_{3}OH(l) \rightarrow B(OC H_{3})_{3}(s) + 3 H_{2}O(l)\)

Key concepts

Limiting Reactant

Understanding the concept of the limiting reactant is crucial in predicting the amount of products formed in a chemical reaction. It's like following a recipe for a sandwich, where the quantity of one ingredient, such as slices of bread, dictates how many sandwiches you can make, regardless of how much ham or cheese you have.

In stoichiometry, the limiting reactant is the substance that is completely consumed first, and thus determines the maximum amount of product that can be formed. To identify the limiting reactant, one must:

• Balance the chemical equation to ensure the proper stoichiometric ratios.
• Convert the masses of the reactants to moles using their molar mass.
• Divide the number of moles by the coefficient in the balanced equation to find the reactant that produces the least amount of product.

In the provided exercise, by following these steps, you can calculate that UO₂ will run out before HF, making UO₂ the limiting reactant for reaction a. Applying this process to other reactions allows you to predict the outcome based on the initial quantities of reactants.

Chemical Equation Balancing

Balancing chemical equations is a fundamental skill in chemistry. Just as in algebra where both sides of the equation must be equal, in chemistry, the number of atoms for each element must be the same on both sides of the reaction.

To balance an equation, follow these steps:

• Write down the number of atoms of each element present in the unbalanced equation.
• Adjust coefficients, which are numbers placed in front of compounds, to obtain the same number of atoms of each element on both sides.
• Ensure that the coefficients are in the lowest possible whole number ratio.

For example, the balanced equation for the first reaction in the exercise is UO₂(s) + 4 HF(aq) → UF₄(aq) + 2 H₂O(l), indicating that one molecule of UO₂ reacts with four molecules of HF to produce one molecule of UF₄ and two molecules of water. Correctly balancing equations is essential for accurately performing stoichiometric calculations and identifying limiting reactants.

Mole-to-Mass Conversion

The mole-to-mass conversion is a cornerstone of stoichiometry, bridging the gap between the microscopic world of atoms and molecules and the macroscopic world that we can measure. To perform a mole-to-mass conversion, one must use the molar mass of a substance, which is the mass of one mole of that substance.

Here's how you can do a mole-to-mass conversion:

• Determine the number of moles you have of a substance.
• Find the molar mass from the periodic table (in grams per mole) by adding up the atomic masses of all atoms in the molecule.
• Multiply the number of moles by the molar mass to get the mass in grams.

In the textbook solution provided, the mass of UO₂ was converted to moles, which was then used to determine the amount of product formed, showing 0.00361 moles of UO₂ yields approximately 0.888 grams of UF₄. Without mole-to-mass conversions, we wouldn't be able to relate measurable quantities of substances to the balanced chemical equations for predictive purposes.