Question

The density of ethanol, C2H5OH, is 0.789 g/mL. How many milliliters of ethanol are needed to produce 18.2 g of CO2 according to the following chemical equation?

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

Answer 1

Answer: 12.1 ml of ethanol is needed

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$    

$\text{Moles of} CO_2=\frac{18.2g}{44g/mol}=0.414moles$  

$C_2H_5OH(l)+3O_2(g)\rightarrow 2CO_2(g)+3H_2O(l)$  

According to stoichiometry :

2 moles of $CO_2$ is produced by = 1 mole of $C_2H_5OH$

Thus 0.414 moles of $CO_2$ is produced by=$\frac{1}{2}\times 0.414=0.207moles$  of $C_2H_5OH$

 Mass of $C_2H_5OH=moles\times {\text {Molar mass}}=0.207moles\times 46.07g/mol=9.54g$

Volume of ethanol = $\frac{\text {Mass of ethanol}}{\text {density of ethanol}}=\frac{9.54g}{0.789g/ml}=12.1ml$

12.1 ml of ethanol is needed to produce 18.2 g of $CO_2$