IDNLearn.com connects you with experts who provide accurate and reliable answers. Explore thousands of verified answers from experts and find the solutions you need, no matter the topic.
Sagot :
To predict the products of the combustion reaction and balance the chemical equation, follow these steps:
1. Identify the Reactants:
The reactants are:
- [tex]\((CH_3)_3CH_4\)[/tex] (a hydrocarbon in gaseous form)
- [tex]\(O_2\)[/tex] (oxygen gas)
2. Predict the Products of Combustion:
Combustion of a hydrocarbon typically produces:
- Carbon dioxide ([tex]\(CO_2\)[/tex])
- Water ([tex]\(H_2O\)[/tex])
3. Write the Unbalanced Equation:
Write the unbalanced chemical equation for the reaction:
[tex]$ (CH_3)_3CH_4 + O_2 \rightarrow CO_2 + H_2O $[/tex]
4. Balance the Chemical Equation:
To balance the chemical equation, ensure that the number of atoms of each element on the reactant side equals the number on the product side. Let's break down the balancing process:
i. Identify the number of each type of atom in the reactants and products:
- In [tex]\((CH_3)_3CH_4\)[/tex], there is a total of 10 Carbon atoms and 22 Hydrogen atoms.
- For [tex]\(O_2\)[/tex], initially, there are 2 Oxygen atoms.
- On the product side, we need to balance Carbon, Hydrogen, and Oxygen atoms appropriately.
ii. Balance Carbon atoms:
To balance the carbons, we need 10 [tex]\(CO_2\)[/tex] molecules because there are 10 Carbon atoms in the reactant hydrocarbon.
[tex]$ (CH_3)_3CH_4 + O_2 \rightarrow 10CO_2 + H_2O $[/tex]
iii. Balance Hydrogen atoms:
To balance the 22 Hydrogen atoms in the reactant hydrocarbon, we need 11 [tex]\(H_2O\)[/tex] molecules because each molecule of [tex]\(H_2O\)[/tex] contains 2 Hydrogen atoms.
[tex]$ (CH_3)_3CH_4 + O_2 \rightarrow 10CO_2 + 11H_2O $[/tex]
iv. Balance Oxygen atoms:
On the product side, each [tex]\(CO_2\)[/tex] molecule has 2 Oxygen atoms and each [tex]\(H_2O\)[/tex] molecule has 1 Oxygen atom:
- From [tex]\(10CO_2\)[/tex]: [tex]\(10 \times 2 = 20\)[/tex] Oxygen atoms
- From [tex]\(11H_2O\)[/tex]: [tex]\(11 \times 1 = 11\)[/tex] Oxygen atoms
- Total Oxygen atoms on the product side: [tex]\(20 + 11 = 31\)[/tex]
On the reactant side, initially [tex]\(O_2\)[/tex] must be adjusted to have 31 Oxygen atoms:
[tex]$ \frac{31}{2} O_2 \rightarrow 15.5 O_2 $[/tex]
To remove the fraction, multiply the entire equation by 2:
[tex]$ 2(CH_3)_3CH_4 + 31O_2 \rightarrow 20CO_2 + 22H_2O $[/tex]
So, the balanced chemical equation is:
[tex]$ 2(CH_3)_3CH_4 + 31O_2 \rightarrow 20CO_2 + 22H_2O $[/tex]
Summary:
The products of the combustion reaction of [tex]\((CH_3)_3CH_4\)[/tex] in the presence of [tex]\(O_2\)[/tex] are carbon dioxide ([tex]\(CO_2\)[/tex]) and water ([tex]\(H_2O\)[/tex]), with the balanced equation being:
[tex]$ 2(CH_3)_3CH_4 + 31O_2 \rightarrow 20CO_2 + 22H_2O $[/tex]
1. Identify the Reactants:
The reactants are:
- [tex]\((CH_3)_3CH_4\)[/tex] (a hydrocarbon in gaseous form)
- [tex]\(O_2\)[/tex] (oxygen gas)
2. Predict the Products of Combustion:
Combustion of a hydrocarbon typically produces:
- Carbon dioxide ([tex]\(CO_2\)[/tex])
- Water ([tex]\(H_2O\)[/tex])
3. Write the Unbalanced Equation:
Write the unbalanced chemical equation for the reaction:
[tex]$ (CH_3)_3CH_4 + O_2 \rightarrow CO_2 + H_2O $[/tex]
4. Balance the Chemical Equation:
To balance the chemical equation, ensure that the number of atoms of each element on the reactant side equals the number on the product side. Let's break down the balancing process:
i. Identify the number of each type of atom in the reactants and products:
- In [tex]\((CH_3)_3CH_4\)[/tex], there is a total of 10 Carbon atoms and 22 Hydrogen atoms.
- For [tex]\(O_2\)[/tex], initially, there are 2 Oxygen atoms.
- On the product side, we need to balance Carbon, Hydrogen, and Oxygen atoms appropriately.
ii. Balance Carbon atoms:
To balance the carbons, we need 10 [tex]\(CO_2\)[/tex] molecules because there are 10 Carbon atoms in the reactant hydrocarbon.
[tex]$ (CH_3)_3CH_4 + O_2 \rightarrow 10CO_2 + H_2O $[/tex]
iii. Balance Hydrogen atoms:
To balance the 22 Hydrogen atoms in the reactant hydrocarbon, we need 11 [tex]\(H_2O\)[/tex] molecules because each molecule of [tex]\(H_2O\)[/tex] contains 2 Hydrogen atoms.
[tex]$ (CH_3)_3CH_4 + O_2 \rightarrow 10CO_2 + 11H_2O $[/tex]
iv. Balance Oxygen atoms:
On the product side, each [tex]\(CO_2\)[/tex] molecule has 2 Oxygen atoms and each [tex]\(H_2O\)[/tex] molecule has 1 Oxygen atom:
- From [tex]\(10CO_2\)[/tex]: [tex]\(10 \times 2 = 20\)[/tex] Oxygen atoms
- From [tex]\(11H_2O\)[/tex]: [tex]\(11 \times 1 = 11\)[/tex] Oxygen atoms
- Total Oxygen atoms on the product side: [tex]\(20 + 11 = 31\)[/tex]
On the reactant side, initially [tex]\(O_2\)[/tex] must be adjusted to have 31 Oxygen atoms:
[tex]$ \frac{31}{2} O_2 \rightarrow 15.5 O_2 $[/tex]
To remove the fraction, multiply the entire equation by 2:
[tex]$ 2(CH_3)_3CH_4 + 31O_2 \rightarrow 20CO_2 + 22H_2O $[/tex]
So, the balanced chemical equation is:
[tex]$ 2(CH_3)_3CH_4 + 31O_2 \rightarrow 20CO_2 + 22H_2O $[/tex]
Summary:
The products of the combustion reaction of [tex]\((CH_3)_3CH_4\)[/tex] in the presence of [tex]\(O_2\)[/tex] are carbon dioxide ([tex]\(CO_2\)[/tex]) and water ([tex]\(H_2O\)[/tex]), with the balanced equation being:
[tex]$ 2(CH_3)_3CH_4 + 31O_2 \rightarrow 20CO_2 + 22H_2O $[/tex]
We appreciate your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. Thank you for choosing IDNLearn.com for your queries. We’re here to provide accurate answers, so visit us again soon.
