Enter your answer in the provided box. the balanced equation for the combustion of ethanol (ethyl alcohol)

10 answers
Question:

Enter your answer in the provided box. the balanced equation for the combustion of ethanol (ethyl alcohol) is: c2h5oh() + 3o2(g) → 2co2(g) + 3h2o(g) how many g of co2 will be produced by the combustion of 4 mol of ethanol? g co2

Answers

The mass of carbon dioxide produced is 352 grams

Explanation:

We are given:

Moles of ethanol = 4 mol

For the given chemical equation:

[tex]C_2H_5OH(g)+3O_2(g)\rightarrow 2CO_2(g)+3H_2O(g)[/tex]

By Stoichiometry of the reaction:

1 mole of ethanol produces 2 moles of carbon dioxide

So, 4 moles of ethanol will produce = [tex]\frac{2}{1}\times 4=8mol[/tex] of carbon dioxide

To calculate the mass of carbon dioxide, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]

Moles of carbon dioxide = 8 moles

Molar mass of carbon dioxide = 44 g/mol

Putting values in above equation:

[tex]8mol=\frac{\text{Mass of carbon dioxide}}{44g/mol}\\\\\text{Mass of carbon dioxide}=(8mol\times 44g/mol)=352g[/tex]

Hence, the mass of carbon dioxide produced is 352 grams

For a: The limiting reactant is ethanol.

For b: The theoretical yield of water is 4.27 grams.

For c: The percentage yield of water is 86.6 %

Explanation:

For a:

For ethanol:

To calculate the mass of ethanol, we use the equation:

[tex]\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}[/tex]     ......(1)

Density of ethanol = 0.789 g/mL

Volume of ethanol = 4.60 mL

Putting values in equation 1, we get:

[tex]0.789g/mL=\frac{\text{Mass of ethanol}}{4.60mL}\\\\\text{Mass of ethanol}=(0.789g/mL\times 4.60mL)=3.63g[/tex]

To calculate the number of moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]     .....(2)

Given mass of ethanol = 3.63 g

Molar mass of ethanol = 46.1 g/mol

Putting values in equation 2, we get:

[tex]\text{Moles of ethanol}=\frac{3.63g}{46.1g/mol}=0.079mol[/tex]

For oxygen gas:

Given mass of oxygen gas = 15.50 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 2, we get:

[tex]\text{Moles of oxygen gas}=\frac{15.50g}{32g/mol}=0.484mol[/tex]

The chemical equation for the combustion of ethanol follows:

[tex]C_2H_5OH+3O_2\rightarrow 2CO_2+3H_2O[/tex]

By Stoichiometry of the reaction:

1 mole of ethanol reacts with 3 moles of oxygen gas

So, 0.079 moles of ethanol will react with = [tex]\frac{3}{1}\times 0.079=0.237mol[/tex] of oxygen gas

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

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

Hence, the limiting reactant is ethanol.

For b:

By Stoichiometry of the reaction:

1 mole of ethanol produces 3 moles of water

So, 0.079 moles of ethanol will produce = [tex]\frac{3}{1}\times 0.079=0.237mol[/tex] of water

Now, calculating the theoretical yield of water by using equation 1:

Molar mass of water = 18 g/mol

Moles of water = 0.237 moles

Putting values in equation 2, we get:

[tex]0.237mol=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=(0.237mol\times 18g/mol)=4.27g[/tex]

Hence, the theoretical yield of water is 4.27 grams.

For c:

Calculating the mass of water, by using equation 1, we get:

Density of water = 1.00 g/mL

Volume of water = 3.70 mL

Putting values in equation 1, we get:

[tex]1.00g/mL=\frac{\text{Mass of water}}{3.70mL}\\\\\text{Mass of water}=(1.00g/mL\times 3.70mL)=3.70g[/tex]

To calculate the percentage yield of water, we use the equation:

[tex]\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100[/tex]

Experimental yield of water = 3.70 g

Theoretical yield of water = 4.27 g

Putting values in above equation, we get:

[tex]\%\text{ yield of water}=\frac{3.70g}{4.27g}\times 100\\\\\% \text{yield of water}=86.6\%[/tex]

Hence, the percentage yield of water is 86.6 %

a. Ethanol.

b. 4.21 mL H2O.

c. 87.8%

Explanation:

Hello,

- For the limiting reactant, the available amount of ethanol is less than the ideal amount of ethanol that would react with the given amount of oxygen, that's why it is the limiting one.

- You will find the calculations in the attached picture; take into account that the molecular mass of water is 18 g/mol and ethanol 46 g/mol.

- The percent yield is allowed to be computed in terms of volume as well, because the relationships are conserved.

Best regards!


[tex]The combustion of liquid ethanol (c2h5oh) produces carbon dioxide and water. after 4.60 ml of ethano[/tex]

A. ΔH∘rxn =  21.9 KJ/mol

B. ΔH∘rxn = 103 KJ/mol

C. C2H5OH + 3O2 → 2CO2 + 3H2O

Explanation:

A.

The standard reaction equation is given as:

aA + bB → cC + dD

Its standard enthalpy is given as:

ΔH∘rxn = cΔH∘f(C) + dΔH∘f(D) − aΔH∘f(A) − bΔH∘f(B)

Reaction given to us is:

H2O(l) + CCl4(l) → COCl2(g) + 2HCl(g)

So, its standard enthalpy will be:

ΔH∘rxn = (1)ΔH∘f(CoCl2 (g)) + (2)ΔH∘f(HCl(g)) − (1)ΔH∘f(H2O(l)) − (1)ΔH∘f(CCl4(l))

using the values from table:

ΔH∘rxn = - 218.8 KJ/mol + (2)(- 92.3 KJ/mol) - (- 285.8 KJ/mol) - (- 139.5 KJ/mol)

ΔH∘rxn =  21.9 KJ/mol

B.

Reaction given to us is:

2A + B ⇌ 2C + 2D

So, its standard enthalpy will be:

ΔH∘rxn = (2)ΔH∘f(C) + (2)ΔH∘f(D) − (2)ΔH∘f(A) − (1)ΔH∘f(B)

using the values from table:

ΔH∘rxn = (2)181 KJ/mol + (2)(- 523 KJ/mol) - (2)(- 225 KJ/mol) - (- 337 KJ/mol)

ΔH∘rxn = 103 KJ/mol

C.

Balanced equation for combustion of ethanol is:

C2H5OH + 3O2 → 2CO2 + 3H2O

Answer-  

C2H5OH + 3O2 → 2CO2 + 3H2O

Solution -  

When ethanol burns in air , it reacts with oxygen to produce carbon dioxide and water as by product

The reaction would be  

Ethanol + Air (O2) > Carbon dioxide + water

We will write the half reaction equation  

First Half reaction equation would be –  

C2 > 2C+2 + 4 e-

O2 + 4e- > 2O2

Adding the above two half reactions we get  

C2H5OH + O2 + 4e-  > CO2 + 4e- + H2O

The balanced equation is  

C2H5OH + 3O2 → 2CO2 + 3H2O

C2H5OH(l) + O2(g) → CO2(g) + ?H2O(l)

The coefficient should be 3

Explanation:

C2H5OH(l) + O2(g) → CO2(g) + ?H2O(l)

The chemical equation is a reaction between ethanol and oxygen . The reaction yield is carbon dioxide and water. Balancing the chemical equation requires one to make the number of atoms on the reactant side be equal to the number of atoms on the product side.

C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)

The hydrogen atom on the left hand side of the equation is 6. Therefore, the number of atom of hydrogen should be 6 on the right hand side. The number of oxygen atom on the left hand side is 7. Therefore, the number of Oxygen atom on the right hand side should  7 . Oxygen atom in carbon dioxide(right hand side) is 4 atoms, therefore, it requires 3 atoms of oxygen to make the  oxygen atoms 7.

Using 3 as the coefficient of water in the equation will make hydrogen 6 and oxygen 3 .Adding the 3 oxygen to the already 4 atoms of oxygen in carbon dioxide will make oxygen 7 atoms

 

C2H6O(ethanol) + 3 O2 = 2 CO2 + 3 H2O

Explanation:

C2H6O+3O2→2CO2(g)+3H2O

Explanation:

Combustion of Ethanol is reaction in which ethanol reacts with oxygen to release carbon dioxide and water molecule.

1. C2H6O+O2→CO2(g)+H2O

Given is the chemical equation of combustion of ethanol but it is not balanced .

To balance it

lets balance carbon first  

Since there are two carbon atoms on LHS, to balance it on RHS we add 2 as coefficient on C02 in RHS.

Then we have equation

2. C2H6O+O2→2CO2(g)+H2O

lets balance hydrogen now  

Since there are six hydrogen atoms on LHS, to balance it on RHS we add 3 as coefficient on H20 in RHS. On RHS there are two hydrogen atoms. When   3 is added as coefficient then number of hydrogen atoms becomes 6 (3*2 = 6)

Then we have equation given below

2. C2H6O+O2→2CO2(g)+3H2O

lets balance oxygen now now  

on RHS

no. of atoms of oxygen on RHS in 2CO2 = 2*2 = 4

no. of atoms of oxygen on RHS in 3H2O = 3*1 = 3

Total no. of atoms of oxygen on RHS = 4+3 = 7

no. of atoms of oxygen on LHS in C2H6O = 1*1 = 1

no. of atoms of oxygen on LHS in O2 = 1*2 = 2

Total no. of atoms of oxygen on LHS = 2+1 = 3

We cannot make changes in C2H6O as atoms of carbon and hydrogen are balanced so we have to balance the atoms of O2 on RHS.

We add coefficient 3 to O2 on LHS which makes no of atoms in O2 to be 6 (3*2 = 6)

Thus total atoms of oxygen on RHS becomes 7( one atom of oxygen in  C2H6O and 6 atoms in O2)

Writing it we have equation given below.

C2H6O+3O2→2CO2(g)+3H2O

This is the solution to the problem.

C2H5OH +3O2 -> 2CO2+3H2O

Complete combustion:
C2H5OH + 3 O2 = 2 CO2 + 3 H2O
Incomplete combustion:
C2H5OH + 2 O2 = 2 CO + 3 H2O or
C2H5OH + O2 = 2 C + 3 H2O

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