Question

0.579 g of a gaseous hydrocarbon on complete combustion with air gave 1.76 g of co2 and 0.90 g of H2O. Show that the data illustrates the law of conservation of mass​

Answer 1

Given:

The mass of gaseous hydrocarbon = 0.579 gm

The mass of CO2 produced = 1.76 gm

The mass of H2O produced = 0.9 gm

To Prove:

That the given data illustrates the law of conservation of mass​.

Calculation:

- The mass of carbon in 44 gm CO2 = 12 gm

⇒ The mass of carbon in 1.76 gm CO2 = (12/44) × 1.76

⇒ The mass of carbon in 1.76 gm CO2 = 0.48 gm

- The mass of hydrogen in 18 gm H2O = 2 gm

⇒ The mass of hydrogen in 0.9 gm H2O = (2/18) × 0.9

⇒ The mass of hydrogen in 0.9 gm H2O = 0.1 gm

- The total mass of carbon and hydrogen in the products, M(p) = 0.48 + 0.1

⇒ M(p) = 0.58 gm

-The total mass of hydrogen and carbon in reactants, M(r) = 0.579

- So we have:

M(p) ≈ M(r)

- Since the mass of hydrocarbon is equal to the mass of hydrogen and carbon in the products, the law of conservation of mass​ is verified.