Respuesta :
The answer is C8H18 = 120.6 and C7H8 = 36.4 g
Hydrocarbons produce carbon dioxide and water on combustion in presence of air.
Write the balanced combustion reaction of a mixture of octane and toluene.
- As both octane and toluene are hydrocarbons , they give carbon dioxide and water as the products.
[tex]C_{8}H_{18} + C_{7}H_{8} + \frac{43}{2} O_{2}[/tex] → [tex]15CO_{2} + 13H_{2}O[/tex]
- Now, as per this balanced equation, when both 1:1 mole ratio of octane and toluene reacts, then 15 moles of CO2 is produced and 13 moles of H2O.
- As per question, moles of CO2 = 1.35× moles of H2O.
- To obtain this ratio, the coefficients are set as-
[tex]1.5C8H18 + C7H8 + 25.5O2[/tex]→[tex]19CO2 + 13H2O+ 4.5H2[/tex]
- We know that 3 moles of the octane react with 2 moles of the toluene. So we can set up fractions and ratios to figure out the masses of each. Thus, 3+2= 5 parts total of the mixture.
- Octane's part is 3/5 for the moles.
[tex]0.6* 201\ g= 120.6\ g\ of\ Octane.[/tex]
[tex]Mass of toluene = 157- 120.6 = 36.4 g[/tex]
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The answer is C8H18 = 120.6 and C7H8 = 36.4 g
Hydrocarbons produce carbon dioxide and water on combustion in presence of air.
- Write the balanced combustion reaction of a mixture of octane and toluene.
- As both octane and toluene are hydrocarbons , they give carbon dioxide and water as the products.
- Now, as per this balanced equation, when both 1:1 mole ratio of octane and toluene reacts, then 15 moles of CO2 is produced and 13 moles of H2O.
- As per the uestion, moles of CO2 = 1.35× moles of H2O.
To obtain this ratio, the coefficients are set as:
We know that 3 moles of the octane react with 2 moles of the toluene. So we can set up fractions and ratios to figure out the masses of each. Thus, 3+2= 5 parts total of the mixture.
Octane's part is 3/5 for the moles.
To learn more about the combustion of hydrocarbons, visit:
https://brainly.com/question/3369193?referrer=searchResults
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