5.80 mol of solid A was placed in a sealed 1.00-L container and allowed

5.80 mol of solid A was placed in a sealed 1.00-L container and allowed to decompose into gaseous B and C. The concentration of B steadily increased until it reached 1.20 M, where it remained constant. A(s) equilibrium B(g) +C(g) Then, the container volume was doubled and equilibrium was re-established. How many moles of A remain? Hint: Start by finding the value of K. Make a table that expresses the initial and final amounts of each species. We can use moles in the table so long as we convert back to concentrations before plugging into the K expression.Fill in the missing values in this table, then calculate Kc. Hint: You can use the stoichiometry of the reaction to determine the changes in A and C.

2 months ago

Solution 1

Guest Guest #1626529
2 months ago


3.40 mol


By the reaction equation, the number of moles of B and C is equal, so their concentration must be equal. To determinate the equilibrium constant, the solids are not put in the expression, because their activity is equal to 1, so:

Kc = [B]¹x[C]¹

Kc = 1.2 x 1.2 = 1.44

By the Le Chatilers principle, a change in the volume will change the pressure in the system and will shift the equilibrium. But Kc only changes with the temperature, so the concentration of B and C remains 1.2 mol/L. To find the number of moles, we multiply the concentration by the new volume (2.00 L)

nB = nC = 1.2 x 2 = 2.4 mol

So, because the stoichiometry is 1 mol of A: 1 mol of B: 1 mol of C, it was consumed 2.4 mol of A, then remains:

nA = 5.80 - 2.40 = 3.40 mol

📚 Related Questions

List all orbitals from 1s through 5s according to increasing energy for multielectron atoms. Rank orbitals from smallest to largest energy. 1s,3p,4p,3s,3d,4s,2p,2s,5s.
Solution 1


The given list of orbitals can be ranked as follow:

1s 2s 2p 3s 3p 4s 3d 4p 5s.


The given list of orbitals can be ranked as follow:

1s 2s 2p 3s 3p 4s 3d 4p 5s.

According to the Aufbau principal in ground state of elements electron first occupy the lower energy level then fill the higher energy levels. We know that there four subshells s, p, d and f. The maximum number of electrons in these subshells can be calculated by following formula:

2 (2l +1 )

and l = 0,1,2,3,....

maximum numbers of electrons in s subshell are,


2 ( 2(0) + 1)


so maximum electrons in s subshell are 2.

maximum numbers of electrons in p subshell are,

l = 1

2 ( 2(1) + 1)

2( 2 + 1


so maximum electrons in p subshell are 6.

maximum numbers of electrons in d subshell are,

l = 2

2 ( 2(2) + 1)

2( 4 + 1)


so maximum electrons in d subshell are 10.

maximum numbers of electrons in f subshell are,

l = 3

2 ( 2(3) + 1)

2( 6 + 1)


so maximum electrons in f subshell are 14.

Electron first fill 1s subshell then 2s subshell and in this way they goes to higher energy levels.

Analysis of 20.0 g of a compound containing only calcium and bromine indicates that 4.00 g of calcium are present. What is the empirical formula of the compound formed? A. Ca7Br2 B. Ca3Br2 C. CaBr2 D. Ca5Br2
Solution 1


d. Ca_5Br_2


Mass of calcium = 4.00 g

Molar mass of calcium = 40.078 g/mol

Moles of calcium = 4.00 / 40.078 moles = 0.9981 moles

Given that the compound only contains calcium and bromine. So,

Mass of bromine in the sample = Total mass - Mass of calcium

Mass of the sample = 20.0 g

Mass of bromine in sample = 20.0 - 4.00 g = 16.0 g

Molar mass of bromine = 79.904 g/mol

Moles of Bromine  = 16.0 / 79.904  = 0.2002 moles

Taking the simplest ratio for Ca and Br as:

0.9981 : 0.2002  = 5 : 1

The empirical formula is = Ca_5Br_2

What is the actual yield of carbon dioxide gas, if 2.0 g of sodium carbonate reacts with 5.0 g sulfuric acid, to form carbon dioxide gas with a 60.2% yield?
Solution 1


0.64 g.


Na2CO3  + H2SO4 --->  Na2SO4 + CO2 + H2O

82.99 g of Na2CO3 reacts with 98.07 g of H2SO4 to give 44 g of CO2

So 2 g of Na2CO3  gives a theoretical yield of  (44 / 82.99) * 2 = 1.06g CO2

If the yield is 60.2% the actual yield is 1.06 * 0.602

= 0.64 g.

What glassware would be the most accurate to analytically transfer 10.00 ml of a solution?
Solution 1


The correct answer is a volumetric pipette.


These elements are used to transfer an exactly known volume of standard or sample solutions.

In the upper part, they have an engraved ring called the line. If the pipette is filled to this line and discharged properly, the volume indicated by the pipette is poured.

They are manufactured in different sizes and may have one or two flush marks (double flush pipettes).

Have a nice day!

Which of the following is true? a.Tardigrades can survive both extreme heat and extreme cold. b.Tardigrades can survive extreme heat, but not extreme cold. c.Tardigrades can survive extreme cold, but not extreme heat. d.Tardigrades cannot survive extreme heat or extreme cold.
Solution 1


The correct option is: a. Tardigrades can survive both extreme heat and extreme cold.


Tardigrade, also called the moss piglet or water bear, is a phylum of the most resilient micro-animals that can survive in extreme environmental conditions like dehydration, extreme pressure and temperature, radiation, and also outer space.

These micro-organisms can survive at an extremely low temperature of 1 K and extremely high temperature of about 420 K.

Which of the following substances would release the most amount of heat when they cool from 50oC to 25oC if you started with equal masses of each? Aluminum, 0.920 J/goC Cast iron, 0.461 J/goC or Lead, 0.126 J/goC
Solution 1




Based on the formula:


By assuming that both the mass and the temperature difference is equal for all the metals, the most amount of heat loss will be experienced by the aluminium since it has the highest heat capacity (more released or absorbed energy per unit of mass by degree Celsius).

Best regards.

Combining 0.322 mol Fe2O3 with excess carbon produced 10.0 g Fe. Fe2O3+3C⟶2Fe+3CO What is the actual yield of iron in moles? actual yield: mol What is the theoretical yield of iron in moles? theoretical yield: mol What is the percent yield? percent yield:
Solution 1

When 0.322 mole of F₂O₃ react with with excess carbon (C), the actual yield in mole, the theoretical yield in mole and the percentage yield are:

1. The actual yield of iron (Fe) in mole is 0.179 mole

2. The theoretical yield of iron (Fe) in mole is 0.644 mole

3. The percentage yield of iron (Fe) is 27.8%

1. Determination of the actual yield in mole.

Mass of Fe = 10 g

Molar mass of Fe = 56 g/mol

Mole of Fe =?

Mole = mass / molar mass

Mole of Fe = 10 / 56

Mole of Fe = 0.179 mole

Thus, the actual yield of Fe in mole is 0.179 mole

2. Determination of the theoretical yield of Fe.

The balanced equation for the reaction is given below:

F₂O₃ + 3C —> 2Fe + 3CO

From the balanced equation above,

1 mole F₂O₃ of reacted to produce 2 moles of Fe.


0.322 mole of F₂O₃ will react to produce = 2 × 0.322 = 0.644 mole of Fe.

Thus, the theoretical yield of Fe in mole is 0.644 mole

3. Determination of the percentage yield

Actual yield of Fe = 0.179 mole

Theoretical yield of Fe = 0.644 mole

Percentage yield =?

Percentage = \frac{Actual}{Theoretical}  * 100\\\\= \frac{0.179}{0.644}  * 100

Percentage yield = 27.8%

Therefore, the percentage yield of Fe is 27.8%

Learn more: brainly.com/question/16947867

Solution 2


Actual yield = 0.1791 moles

Theoretical yield in moles = 0.644 moles

Percent yield = 27.8 %


The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Given: For Fe

Given mass = 10.0 g

Molar mass of Fe = 55.845 g/mol

Moles of Fe = 10.0 g / 55.845 g/mol = 0.1791 moles

So, Actual yield = 0.1791 moles

Given, Moles of Fe_2O_3 = 0.322 moles

According to the given reaction:

Fe_2O_3+3C\rightarrow 2Fe+3CO

1 mole of Fe_2O_3 on reaction produces 2 moles of Fe

0.322 mole of Fe_2O_3 on reaction produces 2*0.322 moles of Fe

Actual moles of Fe produced = 0.644 moles

Theoretical yield in moles = 0.644 moles

Or, Mass of iron = Moles × Molar mass = 0.644 × 55.845 g = 35.96 g

Theoretical yield = 35.96 g

Given experimental yield = 10.0 g

% yield = (Experimental yield / Theoretical yield) × 100 = (10.0/35.96) × 100 = 27.8 %

A mixture of 50.0ml of ammonia gas and 60.0ml of oxygen gas reacts. If all the gases are at the same temperature and pressure, and the reaction continues till one of the gases is completely consumed. What volume of steam is produced if the other product is nitric oxide gas
Solution 1


The volume of steam produced is 72 ml


We need to first write the balanced equation for the reaction between Ammonia and oxygen.

The balanced equation is;

4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g)

Remember the Gay-lussac's law of combining gases;

At constant pressure and temperature, gases react bearing a simple ratio to one another.

In this case;

From the equation the ratio of the reactants and products will be;

4V: 5V: 4V: 6V

We are given, 50 ml of NH₃ and 60 ml of Oxygen

Assuming all the volume of oxygen (60 ml) was completely consumed, then we can use the simple ratios to find the volume of the other reactant consumed and the volume of reactants.

Volume of O₂ = 60 ml (completely consumed)

The volume of Ammonia = ( 60/5)× 4

                                         = 48 ml

Therefore; 48 ml of NH₃ was consumed and 2 ml NH₃ remained

The volume of H₂O (steam produced)

The ratio of O₂ to H₂O is 5: 6


Volume of steam = (60/5) × 6

                             = 72 ml

Therefore; the volume of steam produced is 72 ml

Enter the formula for the compound formed between potassium and sulfur.
Solution 1




A chemical formula is a formula showing the symbols of elements present in a compound.

It shows the symbols of the elements in a compound and the number of atoms of each element in the compound.

To write the formula of the compound we need to;

  • First, identify the elements in the compounds and their symbols
  • In this case, the symbols of potassium and sulfur are K and S respectively.
  • Secondly, we need to know their valency.
  • Potassium has a valency of 1 since it reacts by losing one electron to attain stability while sulfur has a valence of 2 since it reacts by gaining two electrons to attain an octet configuration.
  • So, we have K¹S²
  • Thirdly, we write the formula with each element taking the valency of the other element as a subscript.
  • We get our formula as K₂S.

Prepare ~ 8.75 × 10–4 M CrO4 solution from the ~1.25× 10–3 M K2CrO4 stock solution. Record the exact molarity of the stock solution. Calculate the required amount of stock solution to make 10.00 mL of dilute solution. Adjust this value as necessary (ratio) based on the volumetric flask available.
Solution 1


7 mL


To prepare a solution by dilution of another one, we can use the equation:

C1V1 = C2V2

where C is the concentration, V is the volume, 1 is the initial solution, and 2 the final solution. V represents the total volume. K₂CrO₄ dissolver by:

K₂CrO₄ → 2K⁺ + CrO₄⁻²

So, 1 mol of K₂CrO₄ results in 1 mol of CrO₄⁻², the in a solution of 8.75x10⁻⁴M of CrO₄⁻², it will be 8.75x10⁻⁴M of K₂CrO₄, so the volume of the stock solution will be:

1.25x10⁻³xV1 = 8.75x10⁻⁴x10

V1 = (8.75x10⁻³)/(1.25x10⁻³)

V1 = 7 mL