Question

The combustion of propane may be described by the chemical equationC3H8(g) + 5O₂(g) → 3 CO₂(g) + 4H₂O(g)How many grams of O₂(g) are needed to completely burn 85.1 g C3H8(g)?

Answer 1

They give us the balanced equation of the reaction. We must first find the moles of propane that are burned.

The moles of propane will be found by dividing the given mass by the molar mass of propane. The molar mass of propane is:44.1g/mol. So, the moles of C3H8 will be:

`\begin{gathered} molC_3H_8=givengC_3H_8*(1molC_3H_8)/(MolarMass,gC_3H_8) \\ molC_3H_8=85.1gC_3H_8*(1molC_3H_8)/(44.1gC_3H_8)=1.93molC_3H_8 \end{gathered}`

Now, we find the moles of O2 by the stoichiometry of the reaction. We see that according to the stoichiometric coefficients, the ratio O2 to C3H8 is 5/1, so the moles of O2 will be:

`\begin{gathered} molO_2=givenmolC_3H_8*(5molO_2)/(1molC_3H_8) \\ molO_2=1.93molC_3H_8*(5molO_2)/(1molC_3H_8)=9.65molO_2 \end{gathered}`

Now, we multiply the moles of O2 by the molar mass of oxygen to find the grams required. The molar mass of O2 is 31.9988g/mol. The grams of O2 will be:

`\begin{gathered} gO_2=givenmolO_2*(MolarMass,gO_2)/(1molO_2) \\ gO_2=9.65molO_2*(31.9988gO_2)/(1molO_2)=308.7gO_2 \end{gathered}`

Answer: To complete burn 85.1 g of C3H8 are needed 309 grams of O2