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Xelga [282]
3 years ago
14

How many grams of oxygen are required to react with 17.0 grams of octane (c8h18) in the combustion of octane in gasoline?

Chemistry
1 answer:
Keith_Richards [23]3 years ago
8 0
The balanced equation for the combustion of octane is as follows
2C₈H₁₈ + 25O₂ ---> 16CO₂  + 18H₂O
stoichiometry of C₈H₁₈  to O₂ is 2:25
number of octane moles reacted - 17.0 g / 114.2 g/mol = 0.149 mol 
according to molar ratio 
if 2 mol of octane reacts with 25 mol of O₂
then 0.149 mol of octane reacts with - 25 /2 x 0.149 mol = 1.86 mol of O₂
 mass of O₂  - 1.86 mol x 32 g/mol = 59.5 g
59.5 g of O₂ is required to react with 

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pH is the measure of the hydrogen ion concentration while pOH is of hydroxide ion concentration in the solution. The pH is 0.939 and pOH is 13.061 pOH.

pH is the concentration of the hydrogen ion released or gained by the species in the solution that depicts the acidity and basicity of the solution.

pOH is the concentration of the hydroxide ion in the solution and is dependent on the pH as an increase in pH decreases the pOH and vice versa.

Both HCl and HBr are strong acids and gets ionized 100 % in the solution. If we let 1 L of solution for the acids then the concentration of the hydrogen ion will be 0.100 M.

Since both completely dissociate we would just add the molarities of each of the H+ ions together and then calculate the PH and POH from that :

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Substituting values in the equation above:

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Therefore, pH is 0.939 and pOH is 13.061.

Learn more about pH and pOH here:

brainly.com/question/2947041

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