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)?
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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