2) In an experiment, \( 4.7 \mathrm{~g} \mathrm{Al}, \mathrm{O} \), and excess \( \mathrm{HF}(\mathrm{l}) \) were dissolved in \( 500 \mathrm{~mL} \) of \( 0.12 \mathrm{M} \mathrm{NaOH} \) : If \( 6.45 \mathrm{~g} \) NasAlFs was obtained, then what is the: \( \mathrm{Na}=23 \mathrm{~g} / \mathrm{mol} \) Al \( =27 \mathrm{~g} / \mathrm{mol} \) O \( =16 \mathrm{~g} / \mathrm{mol} \) Fr \( 19 \mathrm{~g} / \mathrm{mol} \) a) Actual yield b) Theoretical yield c) Percent yield for this experiment? \( 25 p \) \[ \mathrm{Al}_{2} \mathrm{O}_{3}(\mathrm{~s})+\mathrm{NaOH}(\mathrm{aq})+\mathrm{HF}(\mathrm{l}) \longrightarrow \mathrm{NaAlF}_{3}(\mathrm{~s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \] a) Actual yield
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We know that the molar mass of Al2O3 is (2*27 + 3*16) = 102 g/mol. Given that we have 4.7 g of Al2O3, the number of moles is 4.7 g / 102 g/mol = 0.046 mol. Show more…
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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?
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