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Sagot :
We have the ethanol combustion reaction to produce CO2 and water. The balanced equation of the reaction is:
[tex]4C_2H_2OH+9O_2\rightarrow8CO_2+6H_2O[/tex]We must first find the moles of ethanol corresponding to 421 grams. To do this we divide the mass by the molar mass of ethanol. The molar mass of ethanol is: 46.07g/mol. The moles of ethanol will be:
[tex]molC_2H_2O=421gC_2H_2O\times\frac{1molC_2H_2O}{46.07gC_2H_2O}=9.14molC_2H_2O[/tex]Now, by the stoichiometry of the reaction, we see that the Oxygen to Ethanol ratio is 9/4. So, the moles of oxygen needed will be:
[tex]molO_2=9.14molC_2H_2O\times\frac{9molO_2}{4molC_2H_2O}=20.6molO_2[/tex]We find the grams of oxygen by multiplying the moles by the molar mass. The molar mass of O2 is 31.9988g/mol.
[tex]gO_2=20.6molO_2\times\frac{31.9988gO_2}{1molO_2}=658gO_2[/tex]Answer: To complete the combustion of 421 g of ethanol are needed 658 grams of oxygen
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