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4 years ago

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Alum is a compound used in a variety of applications including cosmetics, water purification, and as a food additive. It can be

synthesized from aluminum metal, sulfuric acid, water, and potassium hydroxide, as seen in the following equation:
2Al(g) + 2KOH(aq) + 4H2SO4(aq) + 10H2O =====> 2KAl(SO4) + 12(H2O) = 3H2

Using the data below, determine the theoretical and percent yield for this alum synthesis. Note that aluminum is the limiting reactant.

Bottle Mass - 9.8981g

Bottle with Aluminum Pieces - 10.8955g

Final Product with Bottle Mass - 19.0414g

1 answer:

mylen [45]4 years ago

5 0

yield = 52.23 %

Explanation:

We have the following chemical reaction:

2 Al (s) + 2 KOH (aq) + 4 H₂SO₄ (aq) + 10 H₂O → 2 KAl(SO₄)₂·12 (H₂O) (s) + 3 H₂ (g)

mass of aluminium = mass of bottle with aluminium pieces - bottle mass

mass of aluminium = 10.8955 - 9.8981 = 0.9974 g

mass of alum = mass of bottle with final product - bottle mass

mass of alum = 19.0414 - 9.8981 = 9.1433 g

number of moles = mass / molecular weight

number of moles of aluminium = 0.9974 / 27 = 0.03694 moles

number of moles of alum (practical) = 9.1433 / 474 = 0.01929 moles

To calculate the theoretical quantity of alum that should be obtained from 0.03694 moles of aluminium we devise the following reasoning:

if 2 moles of aluminium produce 2 moles of alum

then 0.03694 moles of aluminium produce X moles of alum

X = (0.03694 × 2) / 2 = 0.03694 moles of alum (theoretical)

yield = (practical quantity / theoretical quantity) × 100

yield = (0.01929 / 0.03694) × 100

yield = 52.23 %

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reaction yield

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Ymorist [56]

Answer:

$\mathtt{ t_1-t_2= In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2} \ years}$

Explanation:

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Thus:

$In( \dfrac{N_1}{N_o}) = - \lambda t_1$

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However, Suppose the time elapsed from the initial stage to arrive at the weight of the percentage of $^{235}U$ to be = $t_2$

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$In( \dfrac{N_o}{N_2}) = \lambda t_2$ ---- (2)

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$N_2$ = Number of radioactive atoms of $^{235}U$ relating to 3.0 a/o weight

Now, equating equation (1) and (2) together, we have:

$In( \dfrac{N_o}{N_1}) -In( \dfrac{N_o}{N_2}) = \lambda( t_1-t_2)$

replacing the half-life of $^{235}U$ = $7.13 \times 10^8 \ years$

$In( \dfrac{N_2}{N_1}) = \dfrac{In 2}{7.13 \times 10^9}( t_1-t_2)$ ( since $\lambda = \dfrac{In 2}{t_{1/2}}$ )

∴

$\mathtt{In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2}= t_1-t_2}$

The time elapsed signifies how long the isotopic abundance of 235U equal to 3.0 a/o

Thus, The time elapsed is $\mathtt{ t_1-t_2= In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2} \ years}$

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