Answer to Question #88906 in General Chemistry for Dessire Roldan

Question

The combustion of ethanol (C2H5OH) takes place by the following reaction equation
C2H5OH + 3 O2 —> 2 CO2 + 3 H2O
What is the volume of CO2 gas produced by the combustion of excess ethanol by 23.3 grams of O2 gas at 25 degrees Celsius and 1.25atm?

Answer 1

$$C_2H_5OH +3O_2 \rightarrow 2CO_2+3H_2O$$

$$n(O_2)=\frac{m(O_2)}{M(O_2)}=\frac{23.3g}{32.00\frac{g}{mol}}=0.728 mol$$

According to equation 3 moles of $O_2$ gives 2 moles of $CO_2$

We have 0.728 mol of $O_2$ which give x mol of $CO_2$

Solve the proportion:

$$\frac{3}{0.728}=\frac{2}{x}$$

$$x= 0.485 mol$$

$$n(CO_2)= 0.485 mol$$

Use Ideal Gas Law to find volume of $CO_2$

$$PV=nRT$$

$$V=\frac{nRT}{P}=\frac{0.485 mol\times 0.082\frac{L\times atm}{K\times mol}\times (25+273.15)K}{1.25 atm}= 9.49 L$$