The total volume of gas present (in L) : 14.04 L
Reaction(balanced) :
C₂H₅OH(g) + 3O₂(g) → 2CO₂(g) + 3H₂O(l)
mol C₂H₅OH(MW=46,07 g/mol) :
[tex]\tt \dfrac{1.86}{46.07}=0.04[/tex]
mol O₂(MW=32 g/mol) :
[tex]\tt \dfrac{10}{16}=0.625[/tex]
Limiting reactant :
[tex]\tt \dfrac{0.04}{1}\div \dfrac{0.625}{3}=0.04\div 0.21\rightarrow C_2H_5OH~limiting~reactant[/tex]
mol CO₂ =
[tex]\tt \dfrac{2}{1}\times 0.04=0.08[/tex]
mol O₂(unreacted) :
[tex]\tt 0.625-(3\times 0.04)=0.505[/tex]
Conditions at T 25 ° C and P 1 atm are stated by RTP (Room Temperature and Pressure). Vm(molar volume) in this condition = 24 liters / mol
Total volume of gas :
volume O₂+volume CO₂ =
[tex]\tt 0.505\times 24+0.08\times 24=12.12+1.92=14.04~L[/tex]
