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Consider investigation III. Identify the independent variable, explain the effect of reactant concentration on reaction rate using collision theory, and calculate the percentage purity of aluminum.

Chemistry

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83.1%83.1\%

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5.3 Investigation III

  • 5.3.1 Write down the independent variable for this investigation. The independent variable is the factor that is intentionally changed by the experimenter to observe its effect. In this investigation, the concentration of H2SO4\text{H}_2\text{SO}_4 is varied across the runs. The independent variable is the concentration of H2SO4\text{H}_2\text{SO}_4.

  • 5.3.2 Use the collision theory to explain how the average rate of the reaction is affected in this investigation. In this investigation, the concentration of H2SO4\text{H}_2\text{SO}_4 is increased from Run 1 to Run 3. According to collision theory, increasing the concentration of a reactant means there are more reactant particles per unit volume. This leads to a higher frequency of collisions between the reacting particles. Since more collisions occur per unit time, the frequency of effective collisions (collisions with sufficient energy and correct orientation) also increases, resulting in a faster reaction rate.

  • 5.3.3 Calculate the percentage purity of the aluminium.

  1. Calculate the total volume of H2\text{H}_2 gas produced in RUN 3.

    The reaction time is given as 2.6 minutes. Convert this to seconds.

    Time=2.6 min×60 s/min=156 s\text{Time} = 2.6 \text{ min} \times 60 \text{ s/min} = 156 \text{ s}

    The average rate of H2\text{H}_2 production in RUN 3 is 40 cm3⋅s−140 \text{ cm}^3 \cdot \text{s}^{-1}.

    Total Volume of H2=Average Rate×Time\text{Total Volume of H}_2 = \text{Average Rate} \times \text{Time} Total Volume of H2=40 cm3⋅s−1×156 s=6240 cm3\text{Total Volume of H}_2 = 40 \text{ cm}^3 \cdot \text{s}^{-1} \times 156 \text{ s} = 6240 \text{ cm}^3

  2. Calculate the moles of H2\text{H}_2 gas produced.

    The molar gas volume at 60∘C60^\circ \text{C} is given as 27000 cm3⋅mol−127000 \text{ cm}^3 \cdot \text{mol}^{-1}.

    Moles of H2=Total Volume of H2Molar Gas Volume\text{Moles of H}_2 = \frac{\text{Total Volume of H}_2}{\text{Molar Gas Volume}} Moles of H2=6240 cm327000 cm3⋅mol−1=0.23111... mol\text{Moles of H}_2 = \frac{6240 \text{ cm}^3}{27000 \text{ cm}^3 \cdot \text{mol}^{-1}} = 0.23111... \text{ mol}

  3. Calculate the moles of pure Al reacted.

    From the balanced chemical equation: 2Al(s)+3H2SO4(aq)→Al2(SO4)3(aq)+3H2(g)2\text{Al(s)} + 3\text{H}_2\text{SO}_4\text{(aq)} \rightarrow \text{Al}_2(\text{SO}_4)_3\text{(aq)} + 3\text{H}_2\text{(g)}
    The mole ratio of Al to H2\text{H}_2 is 2:32:3.

    Moles of Al=Moles of H2×2 mol Al3 mol H2\text{Moles of Al} = \text{Moles of H}_2 \times \frac{2 \text{ mol Al}}{3 \text{ mol H}_2} Moles of Al=0.23111... mol×23=0.15407... mol\text{Moles of Al} = 0.23111... \text{ mol} \times \frac{2}{3} = 0.15407... \text{ mol}

  4. Calculate the mass of pure Al reacted.

    The molar mass of Al is approximately 26.98 g⋅mol−126.98 \text{ g} \cdot \text{mol}^{-1}.

    Mass of Al=Moles of Al×Molar Mass of Al\text{Mass of Al} = \text{Moles of Al} \times \text{Molar Mass of Al} Mass of Al=0.15407... mol×26.98 g⋅mol−1=4.1568... g\text{Mass of Al} = 0.15407... \text{ mol} \times 26.98 \text{ g} \cdot \text{mol}^{-1} = 4.1568... \text{ g}

  5. Calculate the percentage purity of the aluminium.

    The initial mass of the impure aluminium powder is 5 g5 \text{ g}.

    Percentage Purity=Mass of Pure AlMass of Impure Al Sample×100%\text{Percentage Purity} = \frac{\text{Mass of Pure Al}}{\text{Mass of Impure Al Sample}} \times 100\% Percentage Purity=4.1568... g5 g×100%=83.136...%\text{Percentage Purity} = \frac{4.1568... \text{ g}}{5 \text{ g}} \times 100\% = 83.136... \%

    Rounding to one decimal place:

    Percentage Purity=83.1%\text{Percentage Purity} = \boxed{83.1\%}
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