Question

The combustion of propane may be described by the chemical equationC3H8(g)+5O2(g)⟶3CO2(g)+4H2O(g) How many grams of O2(g) are needed to completely burn 81.4 g C3H8(g)?

Answer 1

295.31 grams of O₂ are needed to completely burn 81.4 g C₃H₈.

Step-by-step explanation

Chemical equation of the combustion reaction

`C_3H_(8(g))+5O_(2(g))\rightarrow3CO_(2(g))+4H_2O_((g))`

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

`\frac{159.99g*81.4g}{44.1\text{ g}}=\frac{13023.186\text{ g}}{44.1}=295.31\text{ grams}`

295.31 grams of O₂ are needed to completely burn 81.4 g C₃H₈.