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

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?

[tex]moles of O_{2} =\frac{8.7 grams ofC_{2} H_{4}*3 moles of O_{2} }{28grams ofC_{2} H_{4}}[/tex]

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?

[tex]volume=\frac{0.31 moles*22.4 L}{1 mole}[/tex]

volume= 6.944 L

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

Abdulazeez10 Abdulazeez10 · Answer 2 · 2021-12-14T11:40:17+01:00

The volume of O₂ at STP that will react with the given C₂H₄ is 20.8 L

First, we will determine the number of moles of O₂ required.

From the given balanced chemical equation,

C₂H₄ + 3O₂ → 2CO₂ + 2H₂O

This means

1 mole of C₂H₄ is required to react completely with 3 moles of O₂

Now, we will determine the number of moles of C₂H₄ present

Mass of C₂H₄ = 8.7 g

Using the formula

[tex]Number\ of\ moles = \frac{Mass}{Molar\ mass}[/tex]

Molar mass of C₂H₄ = 28.05 g/mol

∴ Number of moles of C₂H₄ = [tex]\frac{8.7}{28.05}[/tex]

Number of moles of C₂H₄ = 0.31016 mole

Since

1 mole of C₂H₄ is required to react completely with 3 moles of O₂

Then,

0.31016 mole of C₂H₄ will react completely with 3 × 0.31016 mole of O₂

3 × 0.31016 = 0.9305 mole

∴ Number of moles of O₂ required is 0.9305 mole

Now, for the volume of O₂ at STP that will react

Since 1 mole of a gas occupies 22.4 L at STP

Then,

0.9305 mole of O₂ will occupy 0.9305 × 22.4 L at STP

0.9305 × 22.4 = 20.8432 L

≅ 20.8 L

Hence, the volume of O₂ at STP that will react with the given C₂H₄ is 20.8 L

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