Respuesta :
Considering the definition of STP conditions and reaction stoichiometry, the volume of excess reactant that remains after the reaction is complete is 4.928 L.
Definition of STP condition
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.
Reaction stoichiometry
In first place, the balanced reaction is:
CH₄ + 2 O₂ → CO₂ + 2 H₂O
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
- CH₄: 1 mole
- O₂: 2 moles
- CO₂: 1 mole
- H₂O: 2 moles
Volume of excess reactant
You know that in a certain reaction at STP, 29.2 L of methane CH₄ is combusted with 63.3 L of oxygen O₂.
Considering the definition of STP, you can apply the following rules of three:
- if by definition of STP conditions 22.4 liters are occupied by 1 mole of methane, 29.2 L are occupied by how many moles of methane?
[tex]amount of moles of methane=\frac{29.2 Lx1 mole}{22.4 L}[/tex]
amount of moles of methane= 1.30 moles
- if by definition of STP conditions 22.4 liters are occupied by 1 mole of oxygen, 63.3 L are occupied by how many moles of oxygen?
[tex]amount of moles of oxygen=\frac{63.3 Lx1 mole}{22.4 L}[/tex]
amount of moles of oxygen= 2.82 moles
Now you can apply the following rule of three to determine the excess reagent: if by stoichiometry 2 moles of O₂ reacts with 1 mole of CH₄, if 2.82 moles of O₂ react, how much moles of CH₄ will be needed?
[tex]moles of CH_{4} =\frac{2.82 moles of O_{2}x 1 moles of CH_{4}}{2 moles of O_{2}}[/tex]
moles of CH₄= 1.41 moles
But 1.41 moles of CH₄ are not available, 1.30 moles are available. Since you have less moles than you need to react with 2.82 moles of O₂, methane CH₄ will be the limiting reagent. And oxygen O₂ is excess reagent.
Now you can apply the following rule of three to the amount of O₂ that reacts when the limiting reactant reacts completely: if by stoichiometry 1 mole of CH₄ reacts with 2 moles of O₂, if 1.30 mole of CH₄ react, how much moles of O₂ will be needed?
[tex]moles of O_{2} =\frac{2 moles of O_{2}x 1.30 moles of CH_{4}}{1 moles of CH_{4}}[/tex]
moles of CH₄= 2.60 moles
So, after the reaction is complete, the moles of excess reactant that remains are calculated as:
2.82 moles - 2.60 moles= 0.22 moles
Finally, considering the definition of STP, you can apply the following rule of three: if by definition of STP conditions 1 mole of O₂ occupies a volume of 22.4 liters, 0.22 moles occupies how much volume?
[tex]volume=\frac{0.22 molesx22.4 L}{1 mole}[/tex]
volume= 4.928 L
In summary, the volume of excess reactant that remains after the reaction is complete is 4.928 L.
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