3g of ethane c2h6 on complete combustion e 8.8g of co2 and5.4g of water show that the result are in accordance with the law of conservation of mass
2C2H6 + 7O2 ==> 4CO2 + 6H2O
3............................8.8..........5.4
g O2 needed = 3 x (7 mols H2O/2 mols C2H6) = 10.5g
Is 3 + 10,5 = 8.8 + 5.4. It isn't so the law of conservation of mass is not obeyed.
To determine if the results are in accordance with the law of conservation of mass, we need to compare the mass of the reactants (ethane) with the mass of the products (CO2 and H2O).
First, calculate the molar mass of ethane (C2H6):
C: 2 atoms x 12.01 g/mol = 24.02 g/mol
H: 6 atoms x 1.01 g/mol = 6.06 g/mol
Total molar mass of ethane = 24.02 g/mol + 6.06 g/mol = 30.08 g/mol
Next, calculate the number of moles of ethane:
3 g ÷ 30.08 g/mol = 0.0997 mol
According to the balanced chemical equation for the complete combustion of ethane:
C2H6 + 7/2 O2 -> 2 CO2 + 3 H2O
It states that for every 1 mole of ethane, we get 2 moles of CO2 and 3 moles of H2O.
Using the mole ratio, calculate the theoretical mass of CO2 produced:
0.0997 mol C2H6 x (2 mol CO2 / 1 mol C2H6) x (44.01 g/mol) = 8.77 g CO2
Using the mole ratio, calculate the theoretical mass of H2O produced:
0.0997 mol C2H6 x (3 mol H2O / 1 mol C2H6) x (18.02 g/mol) = 5.11 g H2O
Comparing the calculated values with the experimental values:
CO2: Experimental value = 8.8 g, Theoretical value = 8.77 g
H2O: Experimental value = 5.4 g, Theoretical value = 5.11 g
The results obtained from the reaction are very close to the theoretical values, which confirms that the law of conservation of mass has been obeyed.