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Chapter 4: Problem 75

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

Expert verified
The percent yield for the experiment is 87.6%

Step by step solution

01

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, \(moles \, of \, Al2O3 = \frac{7.81 \, g}{101.96 \, g/mol} = 0.0766 \, mol\).
02

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, \(moles \, of \, NaOH = 0.141 \, mol/L \times 3.5 \, L = 0.494 \, mol\).
03

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.
04

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, \(theoretical \, yield \, of \, Na3AlF6 = 0.1532 \, mol \times 209.94 \, g/mol = 32.18 \, g\).
05

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, \( percent \, yield = \frac{28.2 \, g}{32.18 \, g} \times 100 = 87.6 \%\).

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Most popular questions from this chapter

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