Determine the mass of carbon(iv)oxide produced on burning 104g of ethane (c2 H2 )
C= 12, h= 1, o=16.
2(2 H2 + 502 4 co2 +2H2 0
12×2+1 ×2 4 ×(12+16 ×2)
26g 176g
Amount = molar mass.
104/26 =4 mols.of ethane
2mol c2 H2 4
4 mol 11=x
x=(4 ×4)/2 =8 mole co2
answer please
To determine the mass of carbon(iv)oxide produced on burning 104g of ethane (C2H6), we first need to understand the balanced chemical equation for the combustion of ethane:
2C2H6 + 7O2 -> 4CO2 + 6H2O
From this equation, we can see that 2 moles of ethane (C2H6) react to produce 4 moles of carbon(iv)oxide (CO2).
Next, we calculate the molar mass of ethane (C2H6):
C: 12.01 g/mol (carbon)
H: 1.01 g/mol (hydrogen)
Molar mass of C2H6 = 2(C) + 6(H)
= 2(12.01) + 6(1.01)
= 24.02 + 6.06
= 30.08 g/mol
Now we can determine the number of moles of ethane in 104g:
Moles of ethane (C2H6) = Mass / Molar mass
= 104g / 30.08 g/mol
= 3.45 mol (approximately)
Since 2 moles of ethane react to produce 4 moles of carbon(iv)oxide, we can use stoichiometry to calculate the number of moles of carbon(iv)oxide produced:
Moles of carbon(iv)oxide (CO2) = (Moles of ethane) * (4 moles of CO2 / 2 moles of C2H6)
= 3.45 mol * (4/2)
= 6.9 mol
Finally, we calculate the mass of carbon(iv)oxide produced:
Molar mass of CO2 = 12.01 g/mol (carbon) + 2(16.00 g/mol) (oxygen)
= 12.01 g/mol + 32.00 g/mol
= 44.01 g/mol
Mass of carbon(iv)oxide (CO2) = Moles of CO2 * Molar mass of CO2
= 6.9 mol * 44.01 g/mol
= 303.03 g
Therefore, the mass of carbon(iv)oxide produced on burning 104g of ethane is approximately 303.03 grams.