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Mekhanik [1.2K]
2 years ago
14

Calculate each of the following quantities:

Chemistry
1 answer:
balandron [24]2 years ago
5 0

The number of moles in 78.9 g of sodium perchlorate is 0.644 moles and 3.876 x 10²³ formula units are particles.

<h3>What are formula units?</h3>

Moles are converted to formula units by multiplying them by 6.02 x 1023, or Avogadro's number (Na). The moles and masses of tiny particles are estimated using this method.

Given that the mass of sodium perchlorate is 78.9 g

The molar mass of sodium perchlorate is 122.44 g/mol

Moles of sodium perchlorate will be

Moles (n) = mass ÷ molar mass

n = 78.9 g  ÷ 122.44

n = 0.644  moles

If 1 mole = 6.02 x 10²³

Then, 0.644 moles = 3.876 x 10²³ particles

Thus, there are 0.644 moles of sodium perchlorate in 78.9 g.

Learn more about formula units, here:

brainly.com/question/15488332

#SPJ4

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Answer:

674.26 g of AlI₃

Explanation:

We'll begin by calculating the theoretical yield of aluminum (Al). This can be obtained as follow:

Percentage yield of Al = 67.8%

Actual yield of Al = 30.25 g

Theoretical yield of Al =?

Percentage yield = Actual yield /Theoretical yield × 100/

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67.8 / 100 = 30.25 / Theoretical yield

0.678 = 30.25 / Theoretical yield

Cross multiply

0.678 × Theoretical yield = 30.25

Divide both side by 0.678

Theoretical yield = 30.25 / 0.678

Theoretical yield of Al = 44.62 g

Next, we shall determine the mass of AlI₃ that reacted and the mass of Al produced from the balanced equation. This can be obtained as follow:

AlI₃(s) + 3K(s) → 3KI(s) + Al(s)

Molar mass of AlI₃ = 27 + (3×127)

= 27 + 381 = 408 g/mol

Mass of AlI₃ from the balanced equation = 1 × 408 = 408 g

Molar mass of Al = 27 g/mol

Mass of Al from the balanced equation = 1 × 27 = 27 g

Summary:

From the balanced equation above,

408 g of AlI₃ reacted to produce 27 g of Al.

Finally, we shall determine the mass of

AlI₃ required to produce 44.62 g of Al. This can be obtained as follow:

From the balanced equation above,

408 g of AlI₃ reacted to produce 27 g of Al.

Therefore, Xg of AlI₃ will react to produce 44.62 g of Al i.e

Xg of AlI₃ = (408 × 44.62)/27

Xg of AlI₃ = 674.26 g

Thus, 674.26 g of AlI₃ is needed for the reaction.

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