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.
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.
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:
You know that in a certain reaction at STP, 29.2 L of methane CHâ‚„ is combusted with 63.3 L of oxygen Oâ‚‚.
Considering the definition of STP, you can apply the following rules of three:
[tex]amount of moles of methane=\frac{29.2 Lx1 mole}{22.4 L}[/tex]
amount of moles of methane= 1.30 moles
[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 Oâ‚‚ reacts with 1 mole of CHâ‚„, if 2.82 moles of Oâ‚‚ react, how much moles of CHâ‚„ 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 CHâ‚„= 1.41 moles
But 1.41 moles of CHâ‚„ are not available, 1.30 moles are available. Since you have less moles than you need to react with 2.82 moles of Oâ‚‚, methane CHâ‚„ will be the limiting reagent. And oxygen Oâ‚‚ is excess reagent.
Now you can apply the following rule of three to the amount of Oâ‚‚ that reacts when the limiting reactant reacts completely: if by stoichiometry 1 mole of CHâ‚„ reacts with 2 moles of Oâ‚‚, if 1.30 mole of CHâ‚„ react, how much moles of Oâ‚‚ 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 CHâ‚„= 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 Oâ‚‚ 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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