Question 10.
m = 0.57 g of CHO
m = 1.1 g of CO2, mol = mass/MW = 1.1/44 = 0.025 mol of C
m = 0.65 g of H2O, mol = masS/MW = 0.65/18 = 0.03611 mol of H2O --> 0.07222 mol of H
then
mass of C = 0.025 *12 = 0.3 g
mass of H = 0.07222 *1 = 0.07222 g
mass of O = 0.57 - 0.3 - 0.07222 = 0.19778 g of O
mol = mass/MW = 0.19778/16 = 0.01236 mol of O
ratios
0.025 /0.01236 = 2
0.07222 /0.01236 = 5.8
ratio is then
C2H6O ; this is most likely ethanol, CH3CH2-OH
The equatio of combustion
C2H6O + O2 = CO2 + H2O
balance
C2H6O + 3O2 = 2CO2 + 3H2O
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