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) C3H8 (g) + 5O2 (g) ⟶3CO2 (g) + 4H2O (g) How many grams of O 2 (g) O2 (g) are needed to completely burn 17.9 g C 3 H 8 (g) ?

Question

Chemistry

Javon Kent

23 June, 19:16

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) C3H8 (g) + 5O2 (g) ⟶3CO2 (g) + 4H2O (g) How many grams of O 2 (g) O2 (g) are needed to completely burn 17.9 g C 3 H 8 (g) ?

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Answers (1)

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Kadyn · Answer 1

65.1g of O2 will be needed

Explanation:

The equation for the reaction is given below:

C3H8 + 5O2 - >3CO2 + 4H2O

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

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

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

From the above calculations, we can easily find the mass of O2 needed to completely burn 17.9 g C3H8. This can be done by the following:

From the equation,

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

Therefore,

17.9g of C3H8 will be burnt by Xg of O2. i. e

Xg of O2 = (17.9x160) / 44 = 65.1g