The reaction between dihydrogen sulfide and sulfur dioxide is outlined below. 2 H2S(g) S02(g) -Y 3 S(s) 2 H20(g) a. Identify the limiting reactant when 3.89 g of dihydrogen sulfide react with 4.11 g of sulfur dioxide. Justify your answer. d. Based on your answer from part (a), determine the maximum mass of sulfur that can be produced in this reaction. c. Ifthe actual yield of sulfur is found to be 4.89 g, find the percent yield in this reaction.

Answers

Answer 1

Answer:

Explanation:

2 H₂S(g) +S0₂(g) =  3 S(s) +  2H₂0(g)

2 x 34 g     64 g        3 x 32 g

68 g of  H₂S reacts with 64 g of S0₂

3.89 g of H₂S reacts with 64 x 3 .89 / 68 g of S0₂

3.89 g of H₂S reacts with 3.66  g of S0₂

S0₂ given is 4.11 g , so it is in excess .

Hence H₂S is limiting reagent .

68 g of  H₂S reacts with  S0₂ to give 96 g of Sulphur

3.89 g of  H₂S reacts with  S0₂ to give 96 x 3.89 / 68 g of Sulphur

3.89 g of  H₂S reacts with  S0₂ to give 96 x 3.89 / 68 g of Sulphur

5.49 g of Sulphur is produced .

Actual yield is 4.89

percentage yield = 4.89 x 100 / 5.49

= 89 % .  

Answer 2

Considering the reaction stoichiometry and the definition of percent yield:

a. H₂S will be the limiting reagent.

b. the maximum mass of sulfur that can be produced in the reaction is 5.49 grams.

c. the percent yield of the reaction is 89%.

The balanced reaction is:

2 H₂S(g) + SO₂(g) → 3 S(s) +  2 H₂O(g)

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

H₂S: 2 molesSO₂: 1  moleS: 3 moles  H₂O: 2 moles

The molar mass of each compound is:

H₂S: 34 g/moleSO₂: 64 g/moleS: 32 g/mole  H₂O: 18 g/mole

So, by reaction stoichiometry, the following amounts of mass of each compound participate in the reaction:

H₂S: 2 moles× 34 g/mole= 68 gramsSO₂: 1 moles× 64 g/mole= 64 gramsS: 3 moles× 32 g/mole= 96 grams H₂O: 2 moles× 18 g/mole= 36 grams

a. Limiting reagent

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 64 grams of SO₂ reacts with 68 grams of H₂S, 4.11 grams of SO₂ react with how much mass of H₂S?

[tex]mass of H_{2} S=\frac{4.11 grams of SO_{2}x 68 grams of H_{2} S }{64 grams of SO_{2}}[/tex]

mass of H₂S= 4.37 grams

But 4.37 moles of H₂S are not available, 3.89 grams are available. Since you have less mass than you need to react with 4.11 grams of SO₂, H₂S will be the limiting reagent.

b. Maximum mass of sulfur

Using the limiting reagent, you can apply the following rule of three: if by stoichiometry 68 grams of H₂S produce 96 grams of S, 3.89 grams of H₂S will produce how much mass of S?

[tex]mass of S=\frac{96 grams of Sx 3.89 grams of H_{2} S }{68 grams of H_{2}S}[/tex]

mass of S= 5.49 grams

So, the maximum mass of sulfur that can be produced in the reaction is 5.49 grams.

c. Percent yield

The amount of product that is obtained when all the limiting reagent reacts.  This is called the theoretical yield of the reaction. That is, the theoretical yield is the maximum amount of product that can be produced in a reaction.

On the other hand, the actual yield is the amount of product actually obtained from a reaction.

The percent yield determines the efficiency of the  reaction, and describes the ratio of the actual yield to the theoretical yield:

[tex]percent yield=\frac{actual yield}{theoretical yield}x100[/tex]

In this case, you know:

actual yield= 4.89 gtheoretical yield= 5.49 g

So, the percent yield can be calculated as:

[tex]percent yield=\frac{4.89 grams}{5.49 grams}x100[/tex]

Solving:

percent yield= 89%

Finally, the percent yield of the reaction is 89%.

Learn more about reaction stoichiometry:

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Answer: [tex]CF_{4} + 2Br_{2} \rightarrow CBr_{4} + 2F_{2}[/tex] is a single replacement reaction.

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Answers

Answer:

4258.82 g of Na₂SO₄

Explanation:

From the question given above, the following data were obtained;

Number of molecules of Na₂SO₄ = 1.8055x10²⁵ molecules.

Number of mole of Na₂SO₄ =?

From Avogadro's hypothesis,

6.02×10²³ molecules = 1 mole

Therefore,

6.02×10²³ molecules = 1 mole of Na₂SO₄

Next, we shall determine the mass of 1 mole of Na₂SO₄. This can be obtained as follow:

1 mole of Na₂SO₄ = (23×2) + 32 + (16×4)

= 46 + 32 + 64

= 142 g

Thus,

6.02×10²³ molecules = 142 g of Na₂SO₄

Finally, we shall determine the mass of Na₂SO₄ that contains 1.8055x10²⁵ molecules. This can be obtained as follow:

6.02×10²³ molecules = 142 g of Na₂SO₄

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1.8055x10²⁵ molecules

= (1.8055x10²⁵ × 142) / 6.02×10²³

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Explanation:

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3 . 7= 21

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Answer:

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Explanation:

Physical Properties
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volume (extensive)
mass (extensive)
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melting point (intensive): the temperature at which a substance melts.

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Answers

Answer:

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Explanation:

Hello!

In this case, according to the given description of how the temperature changes for aluminum in agreement to the loss of heat of 6120.0 J, we can use the following equation:

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Answer:

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Decomposition=AB->A+B; 2 Elements that are bonded separating

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Answer:

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Answers

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Answers

Answer:

The answer is "[tex]\bold{CH_3COO^{-} \ (aq) + H^{+}\ (aq) \longrightarrow CH_3COOH \ (aq)}[/tex]"

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PLEASE HELP Describe at least two advantages and two disadvantages of using hydropower as a
source of energy.
PLS ONLY ANSWER THE FULL QUESTION ON HERE DO NOT SEND A LINK FOR ME TO CLICK!!!!!!!!!!!!!!!!!!!!!!!!!!

Answers

Answer:

Advantage: Hydropower is a fueled by water, so it's a clean fuel source.

Advantage: It is a domestic source of energy in the US.

Disadvantage: Fish populations can be impacted if fish cannot migrate upstream past impoundment dams to spawning grounds or if they cannot migrate downstream to the ocean.

Disadvantage: Hydropower plants can be impacted by drought. When water is not available, the  hydropower plants can't produce electricity.

Explanation:

Here are the atomic masses of hypothetical elements:
X = 13.25 amu
Y = 69.23 amu
Z = 109.34 amu
3.8 moles of X2Y5Z3 is equivalent to how many grams?
Enter your answer to zero decimal places (round to the ones place). Do
not include the units of "g", just the numerical answer.

Answers

Answer:

2663 g

Explanation:

We'll begin by calculating the molar mass Of X₂Y₅Z₃. This can be obtained as follow:

Molar mass of X₂Y₅Z₃ = (13.25×2) + (69.23×5) + (109.34×3)

= 26.5 + 346.15 + 328.02

= 700.67 g/mol

Finally, we shall determine the mass of 3.8 moles of X₂Y₅Z₃. This can be obtained as follow:

Molar mass of X₂Y₅Z₃ = 700.67 g/mol

Mole of X₂Y₅Z₃ = 3.8 moles

Mass of X₂Y₅Z₃ =?

Mass = mole × molar mass

Mass of X₂Y₅Z₃ = 3.8 × 700.67

Mass of X₂Y₅Z₃ = 2663 g

Therefore, the of 3.8 moles of X₂Y₅Z₃ is

2663 g

An organism is living in a rapidly changing environment. Sexual reproduction is advantageous to an organism in this situation because–

sexual reproduction produces more offspring than asexual reproduction within a given time period.

sexual reproduction produces offspring with more variation, some of which may be better adapted to a new environment.

sexual reproduction requires more time to complete, allowing the environment to become more stable.

offspring produced by sexual reproduction grows to maturity more rapidly.

Answers

Answer:

sexual reproduction produces offspring with more variation, some of which may be better adapted to a new environment.

Explanation:

this allows more of the organisms to survive a change in the environment

Answer:

sexual reproduction produces offspring with more variation, some of which may be better adapted to a new environment.

Explanation:

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Answers

I think the answer is 101.2 L

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Answers

Answer:

Generally, the first ionisation energy increases along a period. But there are some exceptions one which is not an exception

Bears stop coming to a river ecosystem where they have been eating many fish each day. The fish the bears eat normally eat smaller fish, which eat plants along the river bottom.

What happens to the ecosystem?


Both the larger and the smaller fish populations grow quickly but then die out because the plant life is insufficient for them all to eat.

The larger fish population will drop first, and the smaller fish population will grow quickly. The plants will die off because too many of the smaller fish are eating them.

The larger fish population explodes at first, and the smaller fish population begins to drop. Eventually, the river runs out of smaller fish so larger fish die out, and the plant population grows.

The smaller fish population begins to eat more plants and to grow. The larger fish have more food to eat so their population is able to grow, too.

Answers

Answer:

The larger fish population explodes at first, and the smaller fish population begins to drop. Eventually, the river runs out of smaller fish so larger fish die out, and the plant population grows.

Explanation:

4NH3 + 502 - 6H20 + 4NO

How many grams of O2 are required to produce 0.3 mol of H20?


4NH3 + 5O2 --> 6H2O + 4NO

How mant grams of NO are produced from 1.55 mol of NH3?


4NH3 + 5O2 --> 6H2O + 4NO

How many grams of NO is produced if 12g of 02 is combined with ammonia? ​

Answers

Explanation:

4NH3 + 502 - 6H20 + 4NO

How many grams of O2 are required to produce 0.3 mol of H20?

4NH3 + 5O2 --> 6H2O + 4NO

How mant grams of NO are produced from 1.55 mol of NH3?

4NH3 + 5O2 --> 6H2O + 4NO

How many grams of NO is produced if 12g of 02 is combined with ammonia?

Describe one way that carbon and silicon are alike.
ASAP

Answers

Answer:

they have the same density

A 35 Liter tank of Oxygen is at 315 K with an internal pressure of 190 atmospheres. How many moles of gas does the tank contain?

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Bjhjjbhjhhjjhjbhuhbh

Calculate the final concentration of each of the following:

2.0 L of a 6.0 M HCl solution is added to water so that the final volume is 6.0 L
--
Water is added to 0.50 L of a 12 M NaOH solution to make a 3.0 L of a diluted NaOH solution

Answers

Answer:

1. 2 M

2. 2 M

Explanation:

1. Determination of the final concentration.

Initial Volume (V₁) = 2 L

Initial concentration (C₁) = 6 M

Final volume (V₂) = 6 L

Final concentration (C₂) =?

The final concentration can be obtained as follow:

C₁V₁ = C₂V₂

6 × 2 = C₂ × 6

12 = C₂ × 6

Divide both side by 6

C₂ = 12 / 6

C₂ = 2 M

Therefore, the final concentration of the solution is 2 M

2. Determination of the final concentration.

Initial Volume (V₁) = 0.5 L

Initial concentration (C₁) = 12 M

Final volume (V₂) = 3 L

Final concentration (C₂) =?

The final concentration can be obtained as follow:

C₁V₁ = C₂V₂

12 × 0.5 = C₂ × 3

6 = C₂ × 3

Divide both side by 3

C₂ = 6 / 3

C₂ = 2 M

Therefore, the final concentration of the solution is 2 M

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