Question

What volume of CO2 gas (carbon dioxide has a density of 1.98 g/L) could be obtained from the combustion of 5.895 grams of C3H7OH?

Answer 1

6.54L

Explanations:

The chemical equation that results from the combustion of C₃H₇OH is given as:

`2C_3H_7OH+9O_2\rightarrow6CO_2+8H_2O`

Determine the moles of C₃H₇OH

`\text{Moles of }C_(3)H_(7)OH=\frac{\text{mass}}{Molar\text{ mass}}`

Mass of C₃H₇OH = 5.895grams

Molar mass of C₃H₇OH = 60.095 g/mol

`\begin{gathered} \text{Moles of }C_(3)H_(7)OH=(5.895)/(60.095) \\ \text{Moles of }C_(3)H_(7)OH=0.09809\text{moles} \end{gathered}`

According to stoichiometry, 2 moles of C₃H₇OH produces 6 moles of CO₂, hence the moles of CO₂ will be 3(0.09809) = 0.2943moles

Determine the mass of CO₂

`\begin{gathered} \text{Mass of CO}_2=0.2943*44.01 \\ \text{Mass of CO}_2=12.951\text{grams} \end{gathered}`

Finally, calculate the volume of the CO₂ gas

`\begin{gathered} \text{Density}=\frac{mass}{\text{volume}} \\ \text{Volume}=\frac{Mass}{\text{Density}} \\ \text{Volume}=\frac{12.951g}{1.98\text{g/L}} \\ \text{Volume}=6.54L \end{gathered}`

Hence the required volume of the CO₂ that could be obtained from the combustion of 5.895 grams of C₃H₇OH is 6.54L