Question

Cryolite, \(\mathrm{Na}_{3} \mathrm{AlF}_{6^{\prime}}\) is an important industrial reagent. It is made by the reaction below. $$\begin{array}{r} \mathrm{Al}_{2} \mathrm{O}_{3}(\mathrm{s})+6 \mathrm{NaOH}(\mathrm{aq})+12 \mathrm{HF}(\mathrm{g}) \longrightarrow 2 \mathrm{Na}_{3} \mathrm{AlF}_{6}(\mathrm{s})+9 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \end{array}$$ In an experiment, \(7.81 \mathrm{g} \mathrm{Al}_{2} \mathrm{O}_{3}\) and excess \(\mathrm{HF}(\mathrm{g})\) were dissolved in 3.50 L of 0.141 M NaOH. If 28.2 g \(\mathrm{Na}_{3} \mathrm{AlF}_{6}\) was obtained, then what is the percent yield for this experiment?

Short answer

The percent yield for the experiment is 87.6%

Step-by-step solution

Convert given quantity of Al2O3 to moles

Start by converting the 7.81 g of Al2O3 to moles, using the molar mass of Al2O3, which is \(101.96 \, g/mol\). So, \(moles \, of \, Al2O3 = \frac{7.81 \, g}{101.96 \, g/mol} = 0.0766 \, mol\).

Convert given quantity of NaOH to moles

Then, calculate the moles of NaOH using the given concentration and volume. The formula for this is Molarity = Moles/Volume. So rearranging the formula, it gives Moles = Molarity x Volume. We need to convert 3.5 L to L, hence it will become 3.5 L. So, \(moles \, of \, NaOH = 0.141 \, mol/L \times 3.5 \, L = 0.494 \, mol\).

Identify the limiting reactant

To identify the limiting reactant, divide the number of moles of each reactant by its respective stoichiometric coefficient from the balanced equation. The substance with the smallest result will be the limiting reactant (LR). For Al2O3, the coefficient in the balanced equation is 1 and for NaOH it is 6. Hence, \( Al2O3 = \frac{0.0766 \, mol}{1} = 0.0766 \) and \( NaOH = \frac{0.494 \, mol}{6} = 0.0823 \). So, Al2O3 is the limiting reactant.

Calculate the theoretical yield of Na3AlF6

Using the stoichiometric ratio from the balanced equation, calculate the theoretical yield of Na3AlF6 that can be produced from the limiting reagent. From the balanced equation, 2 moles of Na3AlF6 can be produced from 1 mole of Al2O3. So, \(theoretical \, yield \, of \, Na3AlF6 = 0.0766 \, mol \times 2 = 0.1532 \, mol\). Then, convert this to grams using the molar mass of Na3AlF6, which is \(209.94 \, g/mol\). Hence, \(theoretical \, yield \, of \, Na3AlF6 = 0.1532 \, mol \times 209.94 \, g/mol = 32.18 \, g\).

Calculate the percent yield for the experiment

Finally, the percent yield is calculated by dividing the actual yield by the theoretical yield and then multiplying the result by 100. The actual yield is the 28.2 g of Na3AlF6 that was actually obtained. Therefore, \( percent \, yield = \frac{28.2 \, g}{32.18 \, g} \times 100 = 87.6 \%\).

Key concepts

Stoichiometry

Stoichiometry is all about the math of chemical reactions. It involves using a balanced chemical equation to figure out how much of each substance is involved in a reaction. In our example, we have the reaction of aluminum oxide (\(\mathrm{Al_2O_3}\)), sodium hydroxide (\(\mathrm{NaOH}\)), and hydrogen fluoride (\(\mathrm{HF}\)) to produce cryolite (\(\mathrm{Na_3AlF_6}\)) and water.
To do stoichiometry, you need a balanced chemical equation. Here, it already is:**\[\mathrm{Al_2O_3(s) + 6 \mathrm{NaOH(aq)} + 12 \mathrm{HF(g)} \rightarrow 2 \mathrm{Na_3AlF_6(s)} + 9 \mathrm{H_2O(l)}\]**.

Using this equation, you can determine the amount of reactants needed or products formed based on the ratios given by their coefficients. This is crucial for understanding how much of your starting materials are needed and how much product you can expect to produce.In this reaction, molar ratios from the coefficients tell us that 1 mole of \(\mathrm{Al_2O_3}\) reacts with 6 moles of \(\mathrm{NaOH}\) and 12 moles of \(\mathrm{HF}\) to make 2 moles of \(\mathrm{Na_3AlF_6}\) and 9 moles of water. This forms the foundational basis upon which we can calculate quantities.

Limiting Reactant

The limiting reactant is a key player in any chemical reaction. It's the substance that gets used up first and essentially limits how much product you can make. Once it is finished, the reaction stops, no matter how much of the other reactants are left.

To determine the limiting reactant, compare the mole ratio of each reactant to its stoichiometric coefficient in the balanced equation. In our exercise:- For \(\mathrm{Al_2O_3}\), the stoichiometric coefficient is 1.- For \(\mathrm{NaOH}\), it is 6.Calculate their respective mole amounts divided by these coefficients:- \(\mathrm{Al_2O_3}: \frac{0.0766 \, \text{mol}}{1} = 0.0766\)- \(\mathrm{NaOH}: \frac{0.494 \, \text{mol}}{6} = 0.0823\)

The smaller value indicates the limiting reactant—in this case, \(\mathrm{Al_2O_3}\). This means that \(\mathrm{Al_2O_3}\) will run out first, capping the extent of the reaction and determining the maximum amount of \(\mathrm{Na_3AlF_6}\) that can be produced.

Theoretical Yield

Theoretical yield is the maximum amount of product that could be formed from a given amount of reactants, according to the balanced equation. This assumes perfect reaction efficiency with no waste or side reactions.

Using our reaction's stoichiometry and knowing that \(\mathrm{Al_2O_3}\) is the limiting reactant, we find the theoretical yield of \(\mathrm{Na_3AlF_6}\) with the following calculations:From the balanced equation, 1 mole of \(\mathrm{Al_2O_3}\) yields 2 moles of \(\mathrm{Na_3AlF_6}\). So with 0.0766 moles of \(\mathrm{Al_2O_3}\), you can make:\[ 0.0766 \, \text{mol} \times 2 = 0.1532 \, \text{mol of } \mathrm{Na_3AlF_6}\]

Convert moles to grams using the molar mass of \(\mathrm{Na_3AlF_6}\) (209.94 g/mol):\[ 0.1532 \, \text{mol} \times 209.94 \, \text{g/mol} = 32.18 \, \text{g of } \mathrm{Na_3AlF_6}\]

This calculated amount is your theoretical yield. It's the ideal case scenario, showing the total possible amount of \(\mathrm{Na_3AlF_6}\) under perfect conditions, which is different from the actual yield you get in practice.