How many grams of oxygen, O2, required for the complete combustion of 36 grams of pentane, C5H12 at standard conditions? C5H12 + 8O2 -> 5Co2 + 6 H2O

128 grams of oxygen, O₂, is required for the complete combustion of 36 grams of pentane, C₅H₁₂ under standard conditions.

Explanation:

The balanced reaction is:

C₅H₁₂ + 8 O₂ ⇒ 5 CO₂ + 6 H₂O

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of each compound participate in the reaction:

Being the molar masses of the compounds:

then, by stoichiometry, the following quantities of mass participate in the reaction:

Then you can apply the following rule of three: if by stoichiometry 72 grams of pentane react with 256 grams of oxygen, then 36 grams of pentane with how much mass of oxygen does it react?

mass of oxygen= 128 grams

128 grams of oxygen, O₂, is required for the complete combustion of 36 grams of pentane, C₅H₁₂ under standard conditions.

i need mutiple choices

consider the reaction below.

 no rx

at 500 k, the reaction is at equilibrium with the following concentrations.

[pci5]= 0.0095 m

[pci3] = 0.020

[ci2] = 0.020 m

what is the equilibrium constant for the given reaction?

0.042

0.42

2.4

24