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Aleksandr [31]
2 years ago
5

Calculate the volume of chlorine molecules produced at room temperature and pressure, when 234g of sodium chloride are electroly

sed. (1 mole of chlorine molecules has a volume of 24 dm³ at room temperature and pressure).​
Chemistry
1 answer:
rewona [7]2 years ago
4 0

The volume of chlorine molecules produced at STP would be 96 dm³.

<h3>Stoichiometric problem</h3>

Sodium chloride ionizes during electrolysis to produce sodium and chlorine ions as follows:

NaCl --- > Na^+ + Cl^-

This means that 1 mole of sodium chloride will produce 1 mole of sodium ion and 1 mole of chlorine ion respectively.

Recall that: mole = mass/molar mass

Hence, 234 g of sodium chloride will give:

                         234/58.44 = 4.00 moles.

Thus, the equivalent number of moles of chlorine produced by 234 g of sodium chloride will be 4 moles.

Recall that:

1 mole of every gas at Standard Temperature and Pressure = 24 Liters.

Hence:

4 moles of chlorine = 4 x 24 = 96 Liters or 96 dm³.

More on stoichiometric problems can be found here: brainly.com/question/14465605

#SPJ1

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If 12.5 grams of strontium hydroxide is reacted with 150 mL of 3.5 M carbonic acid, identify the limiting reactant.
vesna_86 [32]

Answer:

Sr(OH)2

Explanation:

We'll begin by calculating the number of mole of carbonic acid in 150mL of 3.5 M carbonic acid solution. This is illustrated below:

Molarity = 3.5M

Volume = 150mL = 150/1000 = 0.15L

Mole of carbonic acid, H2CO3 =..?

Mole = Molarity x Volume

Mole of carbonic acid, H2CO3 = 3.5 x 0.15 = 0.525 mole.

Next, we shall convert 0.525 mole of carbonic acid, H2CO3 to grams.

Mole of H2CO3 = 0.525 mole

Molar mass of H2CO3 = (2x1) + 12 + (16x3) = 62g/mol.

Mass of H2CO3 =..?

Mass = mole x molar mass

Mass of H2CO3 = 0.525 x 62 = 32.55g

Next, we shall write the balanced equation for the reaction. This is given below:

Sr(OH)2 + H2CO3 → SrCO3 + 2H2O

Next, we shall determine the mass of Sr(OH)2 and H2CO3 that reacted from the balanced equation. This is illustrated below:

Molar mass of Sr(OH)2 = 88 + 2(16 + 1) = 88 + 2(17) = 122g/mol

Mass of Sr(OH)2 from the balanced equation = 1 x 122 = 122g

Molar mass of H2CO3 = (2x1) + 12 + (16x3) = 62g/mol.

Mass of H2CO3 from the balanced equation = 1 x 62 = 62g.

From the balanced equation above, 122g of Sr(OH)2 reacted with 62g of H2CO3.

Finally, we shall determine the limiting reactant as follow:

From the balanced equation above, 122g of Sr(OH)2 reacted with 62g of H2CO3.

Therefore, 12.5g of Sr(OH)2 will react with = (12.5 x 62)/122 = 6.35g.

We can see evidently from the calculations made above that it will take 6.35g out 32.55g of H2CO3 to react with 12.5g of Sr(OH)2. Therefore, Sr(OH)2 is the limiting reactant and H2CO3 is the excess reactant

5 0
3 years ago
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maksim [4K]

Answer:

One extraction: 50%

Two extractions: 75%

Three extractions: 87.5%

Four extractions: 93.75%

Explanation:

The following equation relates the fraction q of the compound left in volume V₁ of phase 1 that is extracted n times with volume V₂.

qⁿ = (V₁/(V₁ + KV₂))ⁿ

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When n = 3, q = (1/2)³ = 1/8, so the fraction removed from phase 1 is (1 - 1/8) = 7/8 or 87.5%.

When n = 4, q = (1/2)⁴ = 1/16, so the fraction removed from phase 1 is (1 - 1/16) = 15/16 or 93.75%.

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Answer:

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