Russ F. answered • 12/06/20
Experienced STEM Educator
What is the total volume of gases at the end of reaction?
Closed rxn vessel: V (of system) is constant
60mL Total of reactants:
20mL CO
40mL O2
2CO + O2 --> 2CO2
Step 1: Determine the limiting reactant.
20mL CO[2 mol CO2/2mol CO] = 20mL CO2
40mL O2[2 mol CO2/1 mol O2] = 80mL CO2
CO is the limiting reactant.
Step 2: Determine volume of non-limiting reactant is consumed in the rxn.
20mL CO [1 mol O2/2 mol CO] = 10mL O2
We see that 20mL CO will react with 10mL O2 to produce 20mL CO2.
Step 3: Analyze
By the end of reaction, the volume of CO is 0 mL; it’s used up.
We also know that 10 mL O2 is used up.
We are left with 20mL CO2 and 30mL O2 (40mL – 10mL).
Conclusion:
Total Volume of gases at end of rxn (stoichiometric values): 20mL CO2 + 30mL O2 = 50mL gases
What we do not know, however, is the volume of the vessel. This will impact the final volume of the gases (but not the ratios/partial pressures), because gases expand to fill the volume of the vessel that they occupy.
Russ F.
My pleasure. Reach out any time!12/06/20
J.R. S.
12/07/20
Russ F.
Consider the problem from the standpoint of the Law of Conservation of Mass. Step 1: Let’s determine the mass of gases before the start of the reaction. 20mL CO [1 mol CO/22.4e3 mL] [28g CO/mol CO] = 0.025 g CO Remember that only 10mL O2 is consumed. 10mL O2 [1 mol O2/22.4e3 mL] [32g O2/mol O2] = 0.014 g O2 Total Mass of reactants = 0.025g CO + 0.014g O2 = 0.039g Step 2: Let’s determine the mass of gases after the reaction. 20mL CO2 [1 mol CO2/22.4e3 mL] [44g CO2/mol CO2] = 0.039g CO2 Note that the mass of CO2 is equal to the mass of reactants used to produce it. 30mL O2 was never used and is present throughout the reaction: 30mL O2 [1 mol O2/22.4e3 mL] [32g O2/mol O2] = 0.042g O2 unused. Therefore: 0.042g O2 + 0.014 g O2 + 0.025 g CO = 0.039g CO2 + 0.042g O2 0.081g = 0.081g While the volume of gases changes, the mass of the gases within the system is preserved.12/07/20
Islam A.
Thank you very much12/06/20