Question

Please help!! i am struggling with this. first correct answer gets brainliest!

Answer 1

Answer: $1.25dm^3$ of unreacted oxygen is left.

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given volume}}{\text{Molar Volume}}$    

$\text{Moles of} CO_2=\frac{3.60dm^3}{22.4dm^3}=0.161moles$

$\text{Moles of} O_2=\frac{7.25dm^3}{22.4dm^3}=0.324moles$

$C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(l)$  

According to stoichiometry :

3 moles of $CO_2$  = 5 moles of $O_2$  

Thus 0.161 moles of $CO_2$ =$\frac{5}{3}\times 0.161=0.268moles$  of $O_2$

moles of $O_2$  left unreacted = (0.324-0.268) = 0.056

Volume of $O_2$ left unreacted = $moles\times {\text {Molar volume}}=0.056mol\times 22.4dm^3/mol=1.25dm^3$

Thus $1.25dm^3$ of unreacted oxygen is left.