Answer:
Explanation:
The combustion reaction of Octane is:
To calculate the mass of CO₂ and H₂O produced, we need to know the mass of octane combusted.
We calculate the mass of Octane from the given volume and density, using the following <em>conversion factors</em>:
Now we<u> convert 1.24 gallons to mL</u>:
- 1.24 gallon *
4693.4 mL
We <u>calculate the mass of Octane</u>:
- 4693.4 mL * 0.703 g/mL = 3.30 g Octane
Now we use the <em>stoichiometric ratios</em> and <em>molecular weights</em> to <u>calculate the mass of CO₂ and H₂O</u>:
- CO₂ ⇒ 3.30 g Octane ÷ 114g/mol *
* 44 g/mol = 10.19 g CO₂
- H₂O ⇒ 3.30 g Octane ÷ 114g/mol *
* 18 g/mol = 4.69 g H₂O
Taking into account the reaction stoichiometry, 340.0 moles of methane are produced when 85.1 moles of carbon dioxide gas react with excess hydrogen gas
<h3>Reaction stoichiometry</h3>
In first place, the balanced reaction is:
CO₂ + 4 H₄ → CH₄ + 2 H₂O
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
- CO₂: 1 mole
- H₄: 4 moles
- CH₄: 1 mole
- H₂O: 2 moles
<h3>Moles of CH₄ formed</h3>
The following rule of three can be applied: if by reaction stoichiometry 1 mole of CO₂ form 4 moles of CH₄, 85.1 moles of CO₂ form how many moles of CH₄?
<u><em>moles of CH₄= 340.4 moles</em></u>
Then, 340.0 moles of methane are produced when 85.1 moles of carbon dioxide gas react with excess hydrogen gas
Learn more about the reaction stoichiometry:
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Answer:
25,050 calories.
Explanation:
A calorie is the amount of energy needed to raise the temperature of one gram of water 1 degree centigrade. If we are raising 835 grams of water 30 degrees then we multiply 835*30 to get 25,050 calories.
