The combustion of propane may be described by the chemical equation C 3 H 8 ( g ) + 5 O 2 ( g ) ⟶ 3 CO 2 ( g ) + 4 H 2 O ( g ) How many grams of O 2 ( g ) are needed to completely burn 80.9 g C 3 H 8 ( g ) ?

294.2g

Explanation:

C3H8 + 5O2 —> 3CO2 + 4H2O

Molar Mass of C3H8 = (3x12) + (8x1) = 36 + 8 = 44g/mol

Molar Mass of O2 = 16x2 = 32g/mol

Mass of O2 from the balanced equation = 5 x 32 = 160g

From the equation,

44g of C3H8 was completely consumed by 160g of O2.

Therefore, 80.9g of C3H8 will be consume by = (80.9x160)/44 = 294.2g of O2

Therefore, 294.2g of O2 is needed