0.44 g of hydrocarbon on complete combustion with oxygen gave 1.8 g of water and 0.88 g of
carbon dioxide. Show that these results are in agreement with the law of conservation of
mass.
Answers
Answer:
Explanation:
=> According to the question, Combustion of hydrocarbon:
Hydrocarbon + Oxygen → Water + Carbon dioxide
=> As per the law of conservation of mass, the total mass of the reactants = the mass of products formed in a chemical reaction.
Thus, Mass of (hydrocarbon + oxygen) = Mass of (Water + Carbon dioxide)
Mass of hydrocarbon = 0.44g
Mass of carbon dioxide = 0.88g
Molar mass of carbon dioxide= 44g
As 44g of carbon dioxide contains 12g of C,
=> Thus, 0.88 g of carbon dioxide will contain = 0.88*12 /44 = 0.24g of C.
0.88×1244 = 0.24g of C
Mass of H₂O = 1.8g
Molar mass of H₂O = 18g
As 18g of water contains 2g of H,
=> Thus, 1.8g of water will contain = 1.8*2/18= 0.2g of H
O₂ is present in excess In combustion reaction, thus, amount of oxygen can be ignored.
Amount of carbon and hydrogen present in hydrocarbon ( reactants) = 0.44g (given)
Amount of Carbon and Hydrogen in product as Carbon dioxide and water = (0.24 + 0.2) = 0.44g
=> Thus, mass of C and H in reactant is equals to the mass of C and H in product. This result follows the aw of conservation of mass.
Explanation:
the above is the correct answer
