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2 years ago

Calculate the theoretical yield and percent yield for this experiment

Chemistry

1 answer:

coldgirl [10]2 years ago

7 0

Answer:

A. Theoretical yield of KCl is 36.49g

B. Percentage yield of KCl is 90.4%

Explanation:

The balanced equation for the reaction is given below:

2KCLO3 —> 2KCl + 3O2

Next, we shall determine the mass KClO3 that decomposed and the mass KCl produce from the balanced equation. This is illustrated below:

Molar mass of KClO3 = 39 + 35.5 + (16x3) = 122.5g/mol

Mass of KClO3 from the balanced equation = 2 x 122.5 = 245g

Molar mass of KCl = 39 + 35.5 = 74.5g/mol

Mass of KCl from the balanced equation = 2 x 74.5 = 149g.

From the balanced equation above,

245g of KClO3 decomposed to produce 149g of KCl.

A. Determination of the theoretical yield of KCl. This is illustrated below:

From the balanced equation above,

245g of KClO3 decomposed to produce 149g of KCl.

Therefore, 60g of KClO3 will decompose to produce = (60 x 149)/245 = 36.49g of KCl.

Therefore, the theoretical yield of KCl is 36.49g

B. Determination of the percentage yield of KCl. This can be obtaine as follow:

Actual yield = 33g

Theoretical yield = 36.49g

Percentage yield =..?

Percentage yield = Actual yield /Theoretical yield x 100

Percentage yield = 33/36.49 x 100

Percentage yield of KCl = 90.4%

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Answer:

A) 10.75 is the concentration of arsenic in the sample in parts per billion .

B) 7,633.66 kg the total mass of arsenic in the lake that the company have to remove.

C) It will take 1.37 years to remove all of the arsenic from the lake.

Explanation:

A) Mass of arsenic in lake water sample = 164.5 ng

The ppb is the amount of solute (in micrograms) present in kilogram of a solvent. It is also known as parts-per million.

To calculate the ppm of oxygen in sea water, we use the equation:

$\text{ppb}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 10^9$

Both the masses are in grams.

We are given:

Mass of arsenic = 164.5 ng = $164.5\times 10^{-9} g$

$1 ng=10^{-9} g$

Volume of the sample = V = $15.3 cm^3$

Density of the lake water sample ,d= $1.00 g/cm^3$

Mass of sample = M = $d\times V=1.0 g/cm^3\times 15.3 cm^3=15.3 g$

$ppb=\frac{164.5\times 10^{-9} g}{15.3 g}\times 10^9=10.75$

10.75 is the concentration of arsenic in the sample in parts per billion.

Mass of arsenic in $1 cm^3$ of lake water = $\frac{164.5\times 10^{-9} g}{15.3}=1.075\times 10^{-8} g$

Mass of arsenic in $0.710 km^3$ lake water be m.

$1 km^3=10^{15} cm^3$

Mass of arsenic in $0.710\times 10^{15} cm^3$ lake water :

$m=0.710\times 10^{15}\times 1.075\times 10^{-8} g=7,633,660.130 g$

1 g = 0.001 kg

7,633,660.130 g = 7,633,660.130 × 0.001 kg=7,633.660130 kg ≈ 7,633.66 kg

7,633.66 kg the total mass of arsenic in the lake that the company have to remove.

Company claims that it takes 2.74 days to remove 41.90 kilogram of arsenic from lake water.

Days required to remove 1 kilogram of arsenic from the lake water :

$\frac{2.74}{41.90} days$

Then days required to remove 7,633.66 kg of arsenic from the lake water :

$=7,633.66\times \frac{2.74}{41.90} days=499.19 days$

1 year = 365 days

499.19 days = $\frac{499.19}{365} years = 1.367 years\approx 1.37 years$

It will take 1.37 years to remove all of the arsenic from the lake.

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