ᴀɴꜱᴡᴇʀ ʙᴏᴛʜ ᴏꜰ ᴛʜᴇ Qᴜᴇꜱᴛɪᴏɴꜱ ...(◍•ᴗ•◍)❤
Answers
Question 3 :
0.22g of a hydrocarbon (i.e., a compound conatining carbon and hydrogen only) on complete combustion with oxygen gave 0.9g water and 0.44g carbon dioxide.
Show that these observations are in accordance with the law of conservation of mass .
Solution :
Here , let us assume that , the hydrocarbon is of the form , C_x H_y .
Hydrocarbons on combustion with oxygen release water and oxygen.
So, the general relation can be written as -
C_x H_y → ( Combustion ) → k H_2O + L CO_2
Now , let us calculate everything with respect to one mole of hydrogen and one mole of oxygen.
So ,
C_x H_y → ( combustion ) → H2O + CO2
So , the hydrocarbon becomes CH2 .
Now, the reaction becomes :
CH2 + O2 → H20 + CO2
According to the law of conservation of mass ,
The masses of the reactants and the products on both the sides of the reaction should be equal .
Number of moles of H2O
=> [ Given mass / Molar mass ]
=> [ 0.9 / 0.18 × 10 ]
=> 1/20 moles or 0.05 moles of H2 molecules are present.
Number of Hydrogen atoms
=> 0.1 Moles
Mass of Hydrogen atoms = 0.1 g .
Now , similarly we can calculate the mass of Carbon atoms.
This will come out to be 0.12 grams .
Therefore the total mass on the products side is 0.22 gram .
Now , this is equal to the mass of the hydrocarbon.
Hence , it is shown that the observations are in accordance with the law of conservation of mass .
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Question 4 :
3 gram of ethane on combustion with 11.2 gram of oxygen gave 8.8 gram of carbon dioxide and 5.4 gram of water . Show that the results are in accordance with the law of conversation of mass .
Solution :
This is a relatively simpler Question .
Total mass of reactants :
=> Mass of Ethane + Mass of Oxygen undergoing Combustion
=> 3g + 11.2 g .
=> 14.2 g .
Total mass of reactants
=> Mass of CO2 released + Mass of O2 released .
=> 8.8 gram + 5.4 gram
=> 14.2 gram .
They are equal .
Hence, the results are in accordance with the law of conservation of mass .
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