Question

Consider the following unbalanced equation:

C2H5OH(g) + O2(g) → CO2(g) + H2O(l)
1.86 g of ethanol reacts with 10.0 g of oxygen. What is the total volume of gas present (in L) after the reaction is complete, assuming the reaction takes place at 1.00 atm and 25°C?

Answer 1

The total volume of gas present (in L) : 14.04 L

Further explanation

Reaction(balanced) :

C₂H₅OH(g) + 3O₂(g) → 2CO₂(g) + 3H₂O(l)

mol C₂H₅OH(MW=46,07 g/mol) :

`\tt (1.86)/(46.07)=0.04`

mol O₂(MW=32 g/mol) :

`\tt (10)/(16)=0.625`

Limiting reactant :

`\tt (0.04)/(1)/ (0.625)/(3)=0.04/ 0.21\rightarrow C_2H_5OH~limiting~reactant`

mol CO₂ =

`\tt (2)/(1)* 0.04=0.08`

mol O₂(unreacted) :

`\tt 0.625-(3* 0.04)=0.505`

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 =

`\tt 0.505* 24+0.08* 24=12.12+1.92=14.04~L`