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2 years ago

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What is the theoretical yield of aluminum that can be produced by the reaction of 60.0 g of aluminum oxide with 30.0 g of carbon

according to the following chemical equation? Al2O3 + 3C → 2Al + 3CO

1 answer:

Ulleksa [173]2 years ago

3 0

Answer:

Theoretical yield = 31.8 g

Explanation:

The formula for the calculation of moles is shown below:

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

For $Al_2O_3$

Mass of $Al_2O_3$ = 60.0 g

Molar mass of $Al_2O_3$ = 101.96128 g/mol

The formula for the calculation of moles is shown below:

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

Thus,

$Moles= \frac{60.0\ g}{101.96128\ g/mol}$

$Moles_{Al_2O_3}= 0.5885\ mol$

Given: For $C$

Given mass = 30.0 g

Molar mass of $C$ = 12.0107 g/mol

The formula for the calculation of moles is shown below:

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

Thus,

$Moles= \frac{30.0\ g}{12.0107\ g/mol}$

$Moles_{C}= 2.4978\ mol$

According to the given reaction:

$Al_2O_3+3C\rightarrow 2Al+3CO$

1 mole of aluminium oxide react with 3 moles of carbon

0.5885 mole of aluminium oxide react with $3\times 0.5885$ moles of carbon

Moles of carbon = 1.7655 moles

Available moles of carbon = 2.4978 moles

Limiting reagent is the one which is present in small amount. Thus, aluminium oxide is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

1 mole of aluminium oxide on reaction forms 2 moles of aluminium.

0.5885 mole of aluminium oxide on reaction forms $2\times 0.5885$ moles of aluminium.

Moles of aluminium = 1.177 moles

Molar mass of aluminium = 26.981539 g/mol

Mass of sodium sulfate = Moles × Molar mass = 1.177 × 26.981539 g = 31.8 g

<u> Theoretical yield = 31.8 g</u>

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How many moles of O2 should be supplied to burn 1 mol of C3H8 (propane) molecules in a camping stove

Shkiper50 [21]

The combustion of any hydrocarbon yields water and carbon dioxide. We will now construct a balanced equation:

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

Each mole of propane requires 5 moles of oxygen.

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2 years ago

The reaction SO2(g)+2H2S(g)←→3S(s)+2H2O(g) is the basis of a suggested method for removal of SO2 from power-plant stack gases. T

kifflom [539]

Answer : The equilibrium $SO_2$ pressure is, $3.93\times 10^{-5}torr$

Explanation :

The given balanced chemical reaction is,

$SO_2(g)+2H_2S(g)\rightleftharpoons 3S(s)+2H_2O(g)$

First we have to calculate the standard free energy of reaction $(\Delta G^o)$ .

$\Delta G^o=G_f_{product}-G_f_{reactant}$

$\Delta G^o=[n_{S(s)}\times \Delta G_f^0_{(S(s))}+n_{H_2O(g)}\times \Delta G_f^0_{(H_2O(g))}]-[n_{SO_2(g)}\times \Delta G_f^0_{(SO_2(g))}+n_{H_2S(g)}\times \Delta G_f^0_{(H_2S(g))}]$

where,

$\Delta G^o$ = standard free energy of reaction = ?

n = number of moles

Now put all the given values in this expression, we get:

$\Delta G^o=[3mole\times (0kJ/mol)+2mole\times (-228.57kJ/mol)]-[1mole\times (-300.4kJ/mol)+2mole\times (-33.01kJ/mol)]$

$\Delta G^o=-90.72kJ/mol$

Now we have to calculate the value of $K_p$

$\Delta G^o=-RT\ln K_p$

where,

$\Delta G_^o$ = standard Gibbs free energy = -90.72 kJ/mol

R = gas constant = 8.314 J/mole.K

T = temperature = 298 K

$K_p$ = equilibrium constant = ?

Now put all the given values in this expression, we get:

$-90.72kJ/mol=-(8.314J/mol.K)\times (298K) \ln K_p$

$K_p=7.98\times 10^{15}$

Now we have to calculate the value of $K_p$ .

The given balanced chemical reaction is,

$SO_2(g)+2H_2S(g)\rightleftharpoons 3S(s)+2H_2O(g)$

The expression for equilibrium constant will be :

$K_p=\frac{(p_{H_2O})^2}{(p_{H_2S})^2\times (p_{SO_2})}$

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

Let the equilibrium $SO_2$ pressure be, x

Pressure of $SO_2$ = Pressure of $H_2S$ = x

Now put all the given values in this expression, we get

$7.98\times 10^{15}=\frac{(22)^2}{(x)^2\times (x)}$

$x=3.93\times 10^{-5}torr$

Thus, the equilibrium $SO_2$ pressure is, $3.93\times 10^{-5}torr$

4 0

2 years ago

There are ________ hydrogen atoms in 25 molecules of c4h4s2.

coldgirl [10]

Answer: 100.

Explanation:

1) The subscripts to the right of each element (symbol) in the chemical formula tells the number of atoms of that element present in one unit formula.

2) The unit formula of C₄H₄S₂ is equal to 1 molecule.

3) Therefore, there are 4 carbon atoms, 4 hydrogen atoms and 2 sulfur atoms in each molecule of C₄H₄S₂.

4) Then, you just have to multiply the corresponding subscript of the element times the number of molecules (25 in this case) to find the number of atoms of that kind.

5) These are the calculations for each element in the molecule C₄H₄S₂.

i) C: 4 × 25 = 100

ii) H: 4 × 25 = 100

iii) S: 2 × 25 = 50.

6) The question is about H only, so the answer is that there are 100 hydrogen atoms in 25 molecules of C₄H₄S₂.

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2 years ago

Read 2 more answers

At 4.00 L, an expandable vessel contains 0.864 mol of oxygen gas. How many liters of oxygen gas must be added at constant temper

fgiga [73]

Answer:

We have to add 2.30 L of oxygen gas

Explanation:

Step 1: Data given

Initial volume = 4.00 L

Number of moles oxygen gas= 0.864 moles

Temperature = constant

Number of moles of oxygen gas increased to 1.36 moles

Step 2: Calculate new volume

V1/n1 = V2/n2

⇒V1 = the initial volume of the vessel = 4.00 L

⇒n1 = the initial number of moles oxygen gas = 0.864 moles

⇒V2 = the nex volume of the vessel

⇒n2 = the increased number of moles oxygen gas = 1.36 moles

4.00L / 0.864 moles = V2 / 1.36 moles

V2 = 6.30 L

The new volume is 6.30 L

Step 3: Calculate the amount of oxygen gas we have to add

6.30 - 4.00 = 2.30 L

We have to add 2.30 L of oxygen gas

4 0

2 years ago

A student reacts 20.0 mL of 0.248 M H2SO4 with 15.0 mL of 0.195 M NaOH. Write a balanced chemical equation to show this reaction

Molodets [167]

<u>Answer:</u> The concentration of salt (sodium sulfate), sulfuric acid and NaOH in the solution is 0.0418 M, 0.0999 M and 0 M respectively.

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ .....(1)

<u>For NaOH:</u>

Initial molarity of NaOH solution = 0.195 M

Volume of solution = 15.0 mL = 0.015 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$0.195M=\frac{\text{Moles of NaOH}}{0.015L}\\\\\text{Moles of NaOH}=(0.195mol/L\times 0.015L)=2.925\times 10^{-3}mol$

<u>For sulfuric acid:</u>

Initial molarity of sulfuric acid solution = 0.248 M

Volume of solution = 20.0 mL = 0.020 L

Putting values in equation 1, we get:

$0.248M=\frac{\text{Moles of }H_2SO_4}{0.020L}\\\\\text{Moles of }H_2SO_4=(0.248mol/L\times 0.020L)=4.96\times 10^{-3}mol$

The chemical equation for the reaction of NaOH and sulfuric acid follows:

$2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O$

By Stoichiometry of the reaction:

2 moles of NaOH reacts with 1 mole of sulfuric acid

So, $2.925\times 10^{-3}$ moles of KOH will react with = $\frac{1}{2}\times 2.925\times 10^{-3}=1.462\times 10^{-3}mol$ of sulfuric acid

As, given amount of sulfuric acid is more than the required amount. So, it is considered as an excess reagent.

Thus, NaOH is considered as a limiting reagent because it limits the formation of product.

Excess moles of sulfuric acid = $(4.96-1.462)\times 10^{-3}=3.498\times 10^{-3}mol$

By Stoichiometry of the reaction:

2 moles of KOH produces 1 mole of sodium sulfate

So, $2.925\times 10^{-3}$ moles of KOH will produce = $\frac{1}{2}\times 2.925\times 10^{-3}=1.462\times 10^{-3}mol$ of sodium sulfate

<u>For sodium sulfate:</u>

Moles of sodium sulfate = $1.462\times 10^{-3}moles$

Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L

Putting values in equation 1, we get:

$\text{Molarity of sodium sulfate}=\frac{1.462\times 10^{-3}}{0.035}=0.0418M$

<u>For sulfuric acid:</u>

Moles of excess sulfuric acid = $3.498\times 10^{-3}mol$

Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L

Putting values in equation 1, we get:

$\text{Molarity of sulfuric acid}=\frac{3.498\times 10^{-3}}{0.035}=0.0999M$

<u>For NaOH:</u>

Moles of NaOH remained = 0 moles

Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L

Putting values in equation 1, we get:

$\text{Molarity of NaOH}=\frac{0}{0.050}=0M$

Hence, the concentration of salt (sodium sulfate), sulfuric acid and NaOH in the solution is 0.0418 M, 0.0999 M and 0 M respectively.

8 0

2 years ago