Question

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

Short answer

The limiting reactant is HF, the mass of the excess reactant (SiO2) is 19.98g, the theoretical yield of \( \mathrm{H}_{2} \mathrm{SiF}_{6} \) is 48.0g, and the percent yield is 95.4%.

Step-by-step solution

Convert the masses of the reactants to moles

\ To determine the limiting reactant, the first thing to do is to convert the masses of the reactants to moles. We use the molar mass of each substance to do this: for SiO2 it is approximately 60.08 g/mol, and for HF it is approximately 20.01 g/mol. For SiO2: \( \frac{40.0 g}{60.08 g/mol} \approx 0.666 mol \) For HF: \( \frac{40.0 g}{20.01 g/mol} \approx 2.00 mol \)

Identify the limiting reactant

\ Now we need to determine which reactant will run out first when the reaction proceeds. Using the stoichiometry of the balanced chemical reaction, we see the ratio of SiO2 to HF is 1:6. This means for every mole of SiO2, the reaction needs 6 moles of HF. But we only have 2 moles of HF, which is not enough to react with all of the SiO2. Therefore, HF is the limiting reactant.

Calculate the mass of the excess reactant

\ Now that we know that HF is the limiting reactant, and it’s all used up, the remaining SiO2 is the excess reactant. We can calculate the mass of SiO2 that didn’t react. From the balanced chemical reaction, 6 mol HF uses 1 mol SiO2. So, 2 mol HF uses \( \frac{2}{6} = 0.333 \) mol SiO2. The remaining SiO2 in moles is \( 0.666 mol - 0.333 mol = 0.333 \) mol. Multiply by the molar mass of SiO2 to convert to grams: \( 0.333 mol * 60.08g/mol = 19.98g \)

Calculate the theoretical yield of \( \mathrm{H}_{2} \mathrm{SiF}_{6} \)

\ The theoretical yield is the calculated amount of product that can be formed from the limiting reactant. From the balanced chemical reaction, 6 mol HF produces 1 mol \( \mathrm{H}_{2} \mathrm{SiF}_{6} \). So, 2 mol HF should produce \( \frac{2}{6} = 0.333 \) mol \( \mathrm{H}_{2} \mathrm{SiF}_{6} \). We convert this to grams using the molar mass of \( \mathrm{H}_{2} \mathrm{SiF}_{6} \), which is 144.09 g/mol. So, theoretical yield = \( 0.333 mol * 144.09 g/mol = 48.0 g \)

Calculate the percent yield

\ The percent yield is calculated by dividing the actual yield by the theoretical yield and multiplying by 100. Percent yield = \( \frac{45.8g}{48.0g} * 100 = 95.4\% \)

Key concepts

Stoichiometry

Stoichiometry is the branch of chemistry that deals with the quantitative relationships of the reactants and products in a chemical reaction. It's the mathematical framework that allows us to predict how much product will be produced from given amounts of reactants and, conversely, how much of the reactants are needed to produce a desired amount of product.

Using the balanced chemical equation as a recipe, stoichiometry involves calculations based on the mole ratios of the reactants and products. These mole ratios (stoichiometric coefficients) are deduced from the balanced equation and provide the key to understanding how reactant quantities relate to each other and to the products formed. For example, in the reaction between SiO₂ and HF, the balanced equation shows a stoichiometric ratio of 1:6, indicating that 1 mole of SiO₂ reacts with 6 moles of HF to produce products.

Understanding stoichiometry is essential for identifying the limiting reactant in a chemical reaction. In our case, identifying HF as the limiting reactant means it dictates the amount of product that can be formed, whereas SiO₂ in excess is left unreacted after the HF is fully consumed. Thus, stoichiometry not only guides the theoretical calculations but also informs practical aspects, like how to economically use materials in industrial processes.

Chemical Reaction Yield

The yield of a chemical reaction refers to the quantity of product obtained from a reaction. There are two types of yield to consider: the theoretical yield and the actual yield. The theoretical yield is the maximum amount of product that can be produced from a given amount of limiting reactant, calculated based on stoichiometry, assuming that the reaction goes to completion and that no product is lost in the process.

In practice, however, the actual yield, which is the amount of product actually obtained from the reaction, is often less than the theoretical yield due to various factors such as incomplete reactions, side reactions, and losses during product recovery. It's crucial for students to differentiate between these two types of yield to accurately assess the efficiency of a reaction.

In our exercise, the theoretical yield of H₂SiF₆ was calculated to be 48.0 g, based on the initial amount of the limiting reactant, HF. However, the actual yield is given as 45.8 g, which is slightly less due to the practical imperfections as mentioned above.

Percent Yield Calculation

Percent yield is an important aspect of chemical reactions, particularly in industrial and laboratory settings, as it measures the efficiency of a reaction. To calculate the percent yield, we use the formula:

Percent yield = \( \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\% \)

In our example, the actual yield of H₂SiF₆ is 45.8 g whereas the theoretical yield is 48.0 g. Inserting these numbers into the percent yield formula gives us:
Percent yield = \( \frac{45.8g}{48.0g} \times 100 = 95.4\% \)

This calculation indicates that the reaction was quite efficient, with the actual yield being 95.4% of the expected theoretical yield. The closer the percent yield is to 100%, the more efficient the reaction. If the percent yield is significantly lower than 100%, it might suggest issues with the reaction or procedure such as the presence of impurities, inaccurate measurements, or incomplete reactions. Understanding how to calculate and interpret percent yield helps students appreciate the practical aspects of chemistry, including the optimization of chemical processes and the need for cost-effective and sustainable chemical production.