Question

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

Answer 1

Answer:

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:

• C: 12 g/mole
• H: 1 g/mole

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?

$moles of O_{2} =\frac{8.7 grams ofC_{2} H_{4}*3 moles of O_{2} }{28grams ofC_{2} H_{4}}$

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?

$volume=\frac{0.31 moles*22.4 L}{1 mole}$

volume= 6.944 L

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