Answer: [tex]1.25dm^3[/tex] of unreacted oxygen is left.
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given volume}}{\text{Molar Volume}}[/tex]
[tex]\text{Moles of} CO_2=\frac{3.60dm^3}{22.4dm^3}=0.161moles[/tex]
[tex]\text{Moles of} O_2=\frac{7.25dm^3}{22.4dm^3}=0.324moles[/tex]
[tex]C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(l)[/tex]
According to stoichiometry :
3 moles of [tex]CO_2[/tex] = 5 moles of [tex]O_2[/tex]
Thus 0.161 moles of [tex]CO_2[/tex] =[tex]\frac{5}{3}\times 0.161=0.268moles[/tex] of [tex]O_2[/tex]
moles of [tex]O_2[/tex] left unreacted = (0.324-0.268) = 0.056
Volume of [tex]O_2[/tex] left unreacted = [tex]moles\times {\text {Molar volume}}=0.056mol\times 22.4dm^3/mol=1.25dm^3[/tex]
Thus [tex]1.25dm^3[/tex] of unreacted oxygen is left.
