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Nostrana [21]
3 years ago
11

A 10.0-g sample of sodium chloride was placed in 10.0 g of water. If 3.85 g of Cl2 was obtained, what was the percent yield of C

l2
Chemistry
1 answer:
suter [353]3 years ago
7 0

Answer:

63.53% yield

Explanation:

The balanced equation for this reaction is 2NaCl + H2O -> 2NaOH +Cl2

First we must find the limiting reactant

From NaCl we can only produce 6.06 grams of Cl2 in <u>theory</u>

From H20 we can only produce 38.995 grams in theory

so we know NaCl is the limiting

% yield is (Actual/Theoretical) x100 so

(3.85/6.06)x100= 63.53% yield

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Calculate the volume of chlorine molecules produced at room temperature and pressure, when 234g of sodium chloride are electroly
rewona [7]

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

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2 years ago
How many molecules of Ca are found in a sample with 0.2 mols?
DochEvi [55]

Answer:

\boxed{1.2 \times 10^{23}\text{ atoms}}

Explanation:

6.023 × 10²³ atoms of Ca are in 1 mol of Ca

\text{No. of atoms} = \text{0.2 mol} \times \dfrac{6.023 \times 10^{23}\text{atoms }}{\text{1 mol }} = \mathbf{1.2 \times 10^{23}} \textbf{ atoms}}\\\\\text{There are }\boxed{\mathbf{1.2 \times 10^{23}} \textbf{ atoms}} \text{ atoms in 0.20 mol of Ca}

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3 years ago
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If the trend is linear, that is boiling point increases by a constant amount for each additional saturated carbon, the boiling point of octane would be 69*2 because you are adding two saturated carbons. So the predicted boiling point is 138C, which is very close to the literature value
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KE = 1/2(8)4 m/s^2
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Kinetic energy would equal 16 J 
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3 years ago
How do you do part B?
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I cannot see the whole equation.Therefore I shall not answer

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