Answer
295.31 grams of O₂ are needed to completely burn 81.4 g C₃H₈.
Explanation
Chemical equation of the combustion reaction
[tex]C_3H_{8(g)}+5O_{2(g)}\rightarrow3CO_{2(g)}+4H_2O_{(g)}[/tex]Molar mass of oxygen = 15.999 g/mol
Molar mass of propane = 44.1 g/mol
From the balanced reaction above, 1 mole propane reacts with 5 moles oxygen to produce 3 and 4 moles of carbon dioxide and water respectively.
⇒ (5 x 2 x 15.999) g = 159.99 g O₂ completely burn 44.1 g C₃H₈,
Therefore, the grams of O₂ needed to completely burn 81.4 g C₃H₈, will be
[tex]\frac{159.99g\times81.4g}{44.1\text{ g}}=\frac{13023.186\text{ g}}{44.1}=295.31\text{ grams}[/tex]295.31 grams of O₂ are needed to completely burn 81.4 g C₃H₈.
