Respuesta :
Answer:
6.54L
Explanations:
The chemical equation that results from the combustion of C₃H₇OH is given as:
[tex]2C_3H_7OH+9O_2\rightarrow6CO_2+8H_2O[/tex]Determine the moles of C₃H₇OH
[tex]\text{Moles of }C_{3}H_{7}OH=\frac{\text{mass}}{Molar\text{ mass}}[/tex]Mass of C₃H₇OH = 5.895grams
Molar mass of C₃H₇OH = 60.095 g/mol
[tex]\begin{gathered} \text{Moles of }C_{3}H_{7}OH=\frac{5.895}{60.095} \\ \text{Moles of }C_{3}H_{7}OH=0.09809\text{moles} \end{gathered}[/tex]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
[tex]\begin{gathered} \text{Mass of CO}_2=0.2943\times44.01 \\ \text{Mass of CO}_2=12.951\text{grams} \end{gathered}[/tex]Finally, calculate the volume of the CO₂ gas
[tex]\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}[/tex]Hence the required volume of the CO₂ that could be obtained from the combustion of 5.895 grams of C₃H₇OH is 6.54L
