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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 60.0 mL of methyl alcohol (d= 0.787 g/mL)?

Respuesta :

transitionstate
Methanol reacts with oxygen as follow,

                           CH₃OH  +  3/2 O₂   →    CO₂  +  2 H₂O
Data Given;

Volume of Methanol  =  60 mL

Density of Methanol  =  0.787 g/mL

Calculating mass of Methanol as;

                          Mass  =  Density × Volume

                          Mass  =  0.787 g/mL × 60 mL

                          Mass  =  47.22 g

According to equation,

                       32 g of Methanol requires  =  48 g (3/2 moles) of O₂

So,

               47.22 g of Methanol will require  =  X g of O₂

Solving for X,
                                   X  =  (47.22 g × 48 g) ÷ 32 g

                                   X  =  70.83 g of O₂

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