Answer to Question #107729 in Chemistry for Katherine Mascorro

Question

How many grams of oxygen are needed to combust 20.0 grams of propane?

Answer 1

propane = C₃H₈.

M(C₃H₈) = 3*Ar(C) + 8*Ar(H) = 3*12 + 8*1 = 44 (g/mol).

Solution:

The equation of a chemical reaction:

C₃H₈ + 5O₂ = 3CO₂ + 4H₂O.

According to the chemical equation: n(C₃H₈) = n(O₂)/5.

n(O₂) = 5 * n(C₃H₈).

n(C₃H₈) = m(C₃H₈) / M(C₃H₈) = (20.0 g) / (44 g/mol) = 0.4545 moles.

Then,

n(O₂) = 5 * n(C₃H₈) = 5 * 0.4545 = 2.2725 moles.

m(O₂) = n(O₂) * M(O₂) = (2.2725 moles) * (32 g/mol) = 72.72 g.

m(O₂) = 72.72 g.

Answer: 72.72 grams of oxygen (O₂) are needed to combust 20.0 grams of propane.