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ICE Princess25 [194]
4 years ago
9

What coefficients are needed to balance the equation for the complete combustion of methane? enter the coefficients in the order

ch4, o2, co2, and h2o, respectively.

Chemistry
2 answers:
Art [367]4 years ago
3 0

The coefficients in the order CH₄, O₂, CO₂, and H₂O, are 1,2,1,2

The equation becomes:

CH₄ (g) + 2O₂ (g) -> CO₂ (g) + 2H₂0 (g)

<h3><em>Further explanation</em></h3>

Complete combustion of hydrocarbon with oxygen will be obtained by CO₂ and H₂O compounds.

If O is insufficient there will be incomplete combustion produced by CO and H and O

Hydrocarbon combustion reactions (specifically alkanes)

\large {\boxed {\bold {C_nH _ (_2_n _ + _ 2_) + \frac {3n + 1} {2} O_2 ----> nCO_2 + (n + 1) H_2O}}}

Equalization of chemical reaction equations can be done using variables.

Steps in equalizing the reaction equation:

  • 1. gives a coefficient on substances involved in the equation of reaction such as a, b, or c etc.
  • 2. make an equation based on the similarity of the number of atoms where the number of atoms = coefficient × index between reactant and product
  • 3. Select the coefficient of the substance with the most complex chemical formula equal to 1

For gas combustion reaction which is a reaction of hydrocarbons with oxygen produces CO₂ and H₂O (water vapor). can use steps:

Balancing C atoms, H and last atoms O atoms

Methana combustion reaction

CH₄ (g) + O₂ (g) -> CO₂ (g) + H₂0 (g)

We give the most complex compounds, namely Methane, we give the number 1

So the reaction becomes

CH₄ (g) + aO₂ (g) -> bCO₂ (g) + cH₂0 (g)

C atom on the left 1, right b, so b = 1

H atom on the left 4, right 2c, so 2c = 4 ---> c = 2

Atom O on the left 2a, right 2b + c, so 2a = 2b + c

2a = 2.1 + 2

2a = 4

a = 2

The equation becomes:

CH₄ (g) + 2O₂ (g) -> CO₂ (g) + 2H₂0 (g)

<h3><em>Learn more</em></h3>

the combustion of octane in gasoline

brainly.com/question/8175791

brainly.com/question/897044

hydrogen and excess oxygen

brainly.com/question/1405182

Keywords: methane, combustion, hydrocarbons, equalization of reaction equations

4vir4ik [10]4 years ago
3 0

The balanced chemical equation is as follows:

{\text{C}}{{\text{H}}_4}\left( g \right) + 2{{\text{O}}_2}\left( g \right) \to {\text{C}}{{\text{O}}_{\text{2}}}\left( g \right) + 2{{\text{H}}_{\text{2}}}{\text{O}}\left( g \right)

The coefficients of {\mathbf{C}}{{\mathbf{H}}_{\mathbf{4}}} is \boxed1, {{\mathbf{O}}_{\mathbf{2}}} is \boxed2, {\mathbf{C}}{{\mathbf{O}}_{\mathbf{2}}} is \boxed1 and {{\mathbf{H}}_{\mathbf{2}}}{\mathbf{O}} is \boxed2

Further Explanation:

The chemical reaction that contains equal number of atoms of the different elements in the reactant as well as in the product side is known as balanced chemical reaction. The chemical equation is required to be balanced to follow the Law of the conservation of mass.

<em>Combustion reaction</em> is the reaction in which the reactant reacts with molecular oxygen to form <em>carbon dioxide</em> and <em>water molecule</em>. <em>Molecular</em> <em>oxygen</em> acts as the <em>oxidizing agent</em> in these reactions. The large amount of heat is released and therefore combustion reactions are <em>exothermic reaction. </em>

The steps to balance a chemical reaction are as follows:

Step 1: Complete the reaction and write the unbalanced symbol equation.

In the combustion reaction, {\text{C}}{{\text{H}}_4} reacts with {{\text{O}}_2} to form {\text{C}}{{\text{O}}_{\text{2}}} and {{\text{H}}_{\text{2}}}{\text{O}}. The physical state of {\text{C}}{{\text{H}}_4} is gas, {{\text{O}}_2} is gas, {\text{C}}{{\text{O}}_{\text{2}}} is gas and {{\text{H}}_{\text{2}}}{\text{O}} is gas. The unbalanced chemical equation is as follows:

{\text{C}}{{\text{H}}_4}\left(g\right)+{{\text{O}}_2}\left(g\right)\to{\text{C}}{{\text{O}}_{\text{2}}}\left(g\right)+{{\text{H}}_{\text{2}}}{\text{O}}\left(g\right)

Step 2: Then we write the number of atoms of all the different elements that are present in a chemical reaction in the reactant side and product side separately. There is one carbon atom on both the reactant and product side. 2 oxygen atoms are present on the reactant side while 3 oxygen atoms are present on the product side. The number of hydrogen atoms on the reactant and product sides is 4 and 2 respectively. (Refer table in the attached image).

Step 3: Initially, we try to balance the number of other atoms of elements except for carbon, oxygen, and hydrogen by multiplying with some number on any side but in the combustion reaction there is only carbon, hydrogen and oxygen atom.

Step 4: After this, we balance the number of atoms of oxygen and then carbon atom followed by hydrogen atoms. To balance the number of atoms of oxygen and hydrogen, we have to multiply {{\text{O}}_2} by 2 and {{\text{H}}_{\text{2}}}{\text{O}} by 2. The chemical equation is as follows:

{\text{C}}{{\text{H}}_4}\left(g\right)+2{{\text{O}}_2}\left(g\right)\to{\text{C}}{{\text{O}}_{\text{2}}}\left(g\right)+2{{\text{H}}_{\text{2}}}{\text{O}}\left(g\right)

Step 5: Finally, we check the number of atoms of each element on both sides. If the number is same then the chemical equation is balanced. The balanced chemical equation is as follows:

{\text{C}}{{\text{H}}_4}\left(g\right)+2{{\text{O}}_2}\left( g \right) \to{\text{C}}{{\text{O}}_{\text{2}}}\left(g\right)+2{{\text{H}}_{\text{2}}}{\text{O}}\left(g\right)

The coefficients of {\text{C}}{{\text{H}}_4} is 1,  {{\text{O}}_2} is 2, {\text{C}}{{\text{O}}_{\text{2}}} is 1 and {{\text{H}}_{\text{2}}}{\text{O}} is 2.

Learn more:

1. Balanced chemical equation brainly.com/question/1405182

2. Learn more about how to calculate moles of the base in given volume brainly.com/question/4283309

Answer details:  

Grade: High School

Subject: Chemistry

Chapter: Chemical reaction and equation

Keywords: Balancing, CH4, O2, CO2, H2O, phases, physical state, solid, liquid, gas, aqueous, coefficients, and combustion reactions.

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B. 2.2

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How many molecules of Oxygen are consumed if 24 molecules of Carbon Dioxide are produced?
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C + O2 → CO2

Mole of C = 24 g/(12 g/mole)

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Mole of molecular O2 = 2.3125 mole

Since mole of C < mole of O2, then C being the limiting reagent.

From the reaction, it shows that mole ratio between C and O2 = 1 : 1.

So, 2 moles of C will stoichiometrically react with 2 moles of O2 to generate 2 moles of CO2.

Avogadro's law states that :"equal volumes of all gases, at the same temperature and pressure, have the same number of molecules i.e. 6.02 x 10^23 molecules/mole.

Therefore, 2 moles of CO2 contain 2 moles x 6.02 x 10^23 molecules/mole = 1.204 x 10^24 molecules of CO2 is formed.

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A sulfuric acid solution containing 571.3 g of h2so4 per liter of aqueous solution has a density of 1.329 g/cm3. Part a calculat
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Mass percentage of a solution is the amount of solute present in 100 g of the solution.

Given data:

Mass of solute H2SO4 = 571.3 g

Volume of the solution = 1 lit = 1000 ml

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Calculations:

Mass of the given volume of solution = 1.329 g * 1000 ml/1 ml = 1329 g

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571.3 g of H2SO4 in 1329 g of the solution

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