Problem 93 - Chemistry Matter and Change

Question

Hydrofluoric acid solutions cannot be stored in glass containers because HF reacts readily with silica dioxide in glass to produce hexafluorosilicic acid \(\left(\mathrm{H}_{2} \mathrm{SiF}_{6}\right)\) \begin{equation} \mathrm{SiO}_{2}(\mathrm{s})+6 \mathrm{HF}(\mathrm{aq}) \rightarrow \mathrm{H}_{2} \mathrm{SiF}_{6}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(1) \end{equation} 40.0 \(\mathrm{g} \mathrm{SiO}_{2}\) and 40.0 \(\mathrm{g}\) HF react to yield 45.8 \(\mathrm{g} \mathrm{H}_{2} \mathrm{SiF}_{6}\) \begin{equation} \begin{array}{l}{\text { a. What is the limiting reactant? }} \\ {\text { b. What is the mass of the excess reactant? }}\end{array} \end{equation} \begin{equation} \begin{array}{l}{\text { c. What is the theoretical yield of } \mathrm{H}_{2} \mathrm{SiF}_{6} \text { ? }} \\ {\text { d. What is the percent yield? }}\end{array} \end{equation}

Step-by-Step Solution

Answer

a. HF is the limiting reactant. b. 20.02 g of SiO2 is in excess. c. The theoretical yield is 48.03 g. d. Percent yield is approximately 95.36%.

Step 1: Calculate Moles of Reactants

First, calculate the moles of \( \mathrm{SiO}_{2} \) and \( \mathrm{HF} \). The molar mass of \( \mathrm{SiO}_{2} \) is approximately 60.08 \text{ g/mol}, so: \[ \text{Moles of } \mathrm{SiO}_{2} = \frac{40.0 \text{ g}}{60.08 \text{ g/mol}} \approx 0.6667 \text{ mol} \] The molar mass of \( \mathrm{HF} \) is approximately 20.01 \text{ g/mol}, so: \[ \text{Moles of } \mathrm{HF} = \frac{40.0 \text{ g}}{20.01 \text{ g/mol}} \approx 2.0005 \text{ mol} \]

Step 2: Identify the Limiting Reactant

From the balanced equation, \(1 \text{ mol of } \mathrm{SiO}_{2} \) reacts with \(6 \text{ mol of } \mathrm{HF} \). The number of moles of \( \mathrm{HF} \) needed: \[ 0.6667 \text{ mol } \mathrm{SiO}_{2} \times 6 = 4.0002 \text{ mol } \mathrm{HF} \] Since there are only 2.0005 moles of \( \mathrm{HF} \), it is the limiting reactant.

Step 3: Calculate Theoretical Yield of Product

Using the moles of the limiting reactant (\( \mathrm{HF} \)), calculate the moles and then the mass of \( \mathrm{H}_{2} \mathrm{SiF}_{6} \). From the equation, \(6 \text{ mol of } \mathrm{HF} \) produces \(1 \text{ mol of } \mathrm{H}_{2} \mathrm{SiF}_{6} \). Thus, \(2.0005 \text{ mol of } \mathrm{HF} \) should produce: \[ \frac{2.0005}{6} \approx 0.3334 \text{ mol } \mathrm{H}_{2} \mathrm{SiF}_{6} \] The molar mass of \( \mathrm{H}_{2} \mathrm{SiF}_{6} \) is approximately 144.09 \text{ g/mol}, so: \[ \text{Theoretical mass } = 0.3334 \text{ mol} \times 144.09 \text{ g/mol} \approx 48.03 \text{ g} \]

Step 4: Calculate Mass of Excess Reactant

\(0.3334 \text{ mol}\) of \( \mathrm{H}_{2} \mathrm{SiF}_{6} \) produced uses \(0.3334 \times 6 = 2.0005 \) mols of \( \mathrm{HF} \). Therefore, \( 0.3334 \times 1 = 0.3334 \) mol of \( \mathrm{SiO}_{2} \) should react, so \(\text{Moles of excess } = 0.6667 - 0.3334 = 0.3333 \text{ mol} \). The mass of the remaining excess reactant: \[ 0.3333 \text{ mol} \times 60.08 \text{ g/mol} \approx 20.02 \text{ g} \]

Step 5: Calculate Percent Yield

Percent yield is given by the formula: \[ \text{Percent Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\% \] Given the actual yield is 45.8 g, calculate: \[ \text{Percent Yield} = \frac{45.8 \text{ g}}{48.03 \text{ g}} \times 100\% \approx 95.36\% \]

Key Concepts

Theoretical Yield

In a chemical reaction, the theoretical yield represents the maximum amount of product that can be generated from the given reactants if everything proceeds perfectly. It is a calculated prediction of what should be produced based on the initial amounts of reactants and the balanced chemical equation. Theoretical yield helps chemists understand the potential output from a chemical process, assuming no side reactions or losses. To calculate the theoretical yield, you first need to determine which reactant is the limiting one. The limiting reactant determines how much product can be formed because it is completely consumed during the reaction, leaving no more material to react further. Once this is determined, you use the stoichiometry of the balanced equation to calculate the moles of product formed from the limiting reactant. Then, by converting these product moles to mass using its molar mass, you obtain the theoretical yield. It's important to remember that the theoretical yield is a perfect scenario calculation and rarely achieved in practice due to losses and inefficiencies in real reactions.

Percent Yield

The percent yield is a crucial metric used to measure the efficiency of a chemical reaction. It compares the actual amount of product obtained (known as the actual yield) with the theoretical yield. This percentage tells chemists how effectively a reaction converted reactants into products. To determine the percent yield, use the following formula:

• Divide the actual yield by the theoretical yield.
• Multiply the result by 100 to convert it into a percentage.

For example, if the theoretical yield of a reaction is 48.03 grams and the reaction actually produces 45.8 grams, the percent yield is:\[\text{Percent Yield} = \left ( \frac{45.8 \text{ g}}{48.03 \text{ g}} \right ) \times 100\% \approx 95.36\%\]Percent yields over 90% are typically considered very good, while anything lower may indicate possible inefficiencies or issues in the reaction process. Keeping track of percent yield helps in the refinement and improvement of chemical processes.

Chemical Reactions

A chemical reaction is a process that transforms one or more substances into new substances with different properties. This transformation involves the breaking and forming of chemical bonds, which rearranges atoms and modifies molecular structures. Chemical reactions can be represented by balanced equations, where reactants are shown on the left and products on the right. The equation must be balanced to respect the law of conservation of mass, meaning the number and type of atoms on each side of the equation must be equal.

Types of Chemical Reactions

There are several main types of chemical reactions, including:

• Combination or synthesis reactions: Two or more substances combine to form a single product.
• Decomposition reactions: A single substance breaks down into two or more products.
• Single displacement reactions: An element replaces a part of a compound, releasing a different element.
• Double displacement reactions: Elements or groups exchange places between two compounds.

Understanding chemical reactions is key in fields such as chemistry, biology, and environmental science. Chemical equations allow chemists to predict the quantities of materials consumed and produced, guiding decisions on scaling lab reactions to industrial level.