The combustion of propane may be described by the chemical equation C3H8(g)+5O2(g)⟶3CO2(g)+4H2O(g) How many grams of O2(g) are needed to completely burn 56.3 g C3H8(g)?

204.7 g

Explanation:

(taking the atomic mass of C, H, O as 12, 1 and 16 respectively).

no. of moles of C3H8 burnt =  56.3 / (12x3 + 1x8)

                                              = 1.27955 mol

From the equation, the mole ratio of C3H8 :  O2 = 1:5

Hence,

the no. of moles of O2 required will be

=1.27955 x 5

= 6.397727 mol

Mass of O2 required = 6.397727 x (16x2)

= 204.7 g

the answer is definitely c.