When burning a gaseous hydrocarbon with a hydrogen density of 21, 8.4 liters of carbon monoxide

When burning a gaseous hydrocarbon with a hydrogen density of 21, 8.4 liters of carbon monoxide (IV) and 6.75 g of water were obtained. Determine the hydrocarbon formula.

1.Let’s find the amount of substance carbon dioxide.

n = V: Vn, where Vn is the molar volume of the gas equal to 22.4 l / mol.

n (CO2) = 8.4 L: 22.4 L / mol = 0.375 mol.

2.Let’s find the mass of carbon dioxide.

m = n M, М (СО2) = 44 g / mol.

m = 0.375 mol × 44 g / mol = 16.5 g.

3.Let’s find the mass of carbon atoms in 16.5 g of carbon dioxide.

44 g – 12 g (C),

16.5 g-m (C).

m = (16.5 × 12): 44,

m = 4.5 g.

4. Let’s find the mass of hydrogen atoms in 6.75 g of water.

Мr (Н2О) = 18 g / mol.

18 g – 2 g (H),

6.75 g-m (H).

m = (6.75 × 2): 18,

m = 0.75 g.

5.Let’s find the amount of matter of carbon and hydrogen atoms.

n = m: M.

M (C) = 12 g / mol.

n (C) = 4.5 g: 12 g / mol = 0.375 mol.

M (H) = 0.75: 1 = 0.75 mol.

6.Let’s find the ratio of the amounts of matter carbon and hydrogen.

C: H = 0.375: 0.75 = 1: 2.

7.Let’s find the molar mass of the hydrocarbon by the relative density.

Relative density is given as hydrogen. You need to multiply the molar mass of hydrogen (2) by 21.

M (H2) = 2 g / mol.

D (H2) = 2 × 21 = 42.

8.Molar mass of hydrocarbon 42. Let’s define the formula.

С3Н6 – propene.

M (C3H6) = 42 g / mol.

Answer: C3H6 is propene.