How many liters of O2 at STP will react with 8.7 grams of C2H4 to form CO2 and H2O? C2H4 + 3 O2 > 2 CO2 + 2 H2O

6.944 liters of O₂ at STP will react with 8.7 grams of C₂H₄ to form CO₂ and H₂O.

Explanation:

The balanced reaction is:

C₂H₄ + 3 O₂ ⇒ 2 CO₂ + 2 H₂O

Being the molar mass of the elements:

Then the molar mass of compound C₂H₄ is:

C₂H₄= 2*12 g/mole + 4*1 g/mole= 28 g/mole

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), 1 mole of C₂H₄ acts. Then, being the mass of a mole of a substance, which can be an element or a compound, 1 mole of C₂H₄ is 28 g, which is the amount of mass that reacts in this case.

Then be able to apply the following rule of three: if by stoichiometry 28 grams of C₂H₄ react with 3 moles of O₂, 8.7 grams of C₂H₄ with how many moles of O₂ do they react?

moles of O₂= 0.31  

The STP conditions refer to the standard temperature and pressure. Pressure values at 1 atmosphere and temperature at 0 ° C are used and are reference values for gases. And in these conditions 1 mole of any gas occupies an approximate volume of 22.4 liters.

Then we can apply the following rule of three: if by definition of STP 1 mole occupies 22.4 L, 0.31 moles how much volume will it occupy?

volume= 6.944 L

6.944 liters of O₂ at STP will react with 8.7 grams of C₂H₄ to form CO₂ and H₂O.

in 1783 cavendish was working with his hydrogen gas and found that when he burned it and condensed the vapor it was water. this was very significant since it disproved the greek theory of elements completely since water was not a simple substance but the product of two gases. it was lavoisier who hearing of the experiment gave cavendish's gas the name hydrogen ("water-producer") and added that hydrogen burned by combining with oxygen and that water was a hydrogen-oxygen combination.

explanation: