Question

Consider the following unbalanced equation:

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) :

$\tt \dfrac{1.86}{46.07}=0.04$

mol O₂(MW=32 g/mol) :

$\tt \dfrac{10}{16}=0.625$

Limiting reactant :

$\tt \dfrac{0.04}{1}\div \dfrac{0.625}{3}=0.04\div 0.21\rightarrow C_2H_5OH~limiting~reactant$

mol CO₂ =

$\tt \dfrac{2}{1}\times 0.04=0.08$

mol O₂(unreacted) :

$\tt 0.625-(3\times 0.04)=0.505$

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 =

$\tt 0.505\times 24+0.08\times 24=12.12+1.92=14.04~L$