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.

explanation:

i did this but i forgot how to do it

since the atomic number is the number of protons in the nucleus, as we move to the left on the periodic table, the electrostatic pull between the nucleus and the valence electron shell decreases. as we move down a group (vertical column), the valence shell energy level gets further and further from the nucleus.