Question

The methyl alcohol (CH3OH) used in alcohol burners combines with oxygen gas to form carbon dioxide...

The methyl alcohol (CH3OH) used in alcohol burners combines with oxygen gas to form carbon dioxide and water. How many grams of oxygen are required to burn 60.0mL of methyl alcohol (d=0.787 g/mL)?

Homework Answers

Answer #1

The combustion reaction of CH3OH is

CH3OH + (3/2) O2 ------------> CO2 + 2H2O

mass of CH3OH = volume x density =  60.0 mL x 0.787 g/mL = 47.22 g

molar mass of methanol (CH3OH) = 32 g/mol

molar mass of oxygen (O2) = 32 g/mol

  CH3OH + (3/2) O2 ------------> CO2 + 2H2O

1 mol = 32 g 1.5 mol = 1.5 x 32 g = 48 g

47.22 g ?

? = (47.22 g/ 32 g) x 48 g O2

= 70.83 g O2

Mass of oxygen required = 70.83 g O2

Therefore,

70.83 grams of oxygen are required to burn 60.0mL of methyl alcohol .

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