How many moles of oxygen are required to burn 34.2 g of methyl alcohol?

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


How many moles of oxygen are required to burn 34.2 g of methyl alcohol?





Our teacher requires us to show work(step 0: balanced equation, 1: grams-moles, 2: mole ratio and 3: moles: to grams so plz. include steps if you can. thanks

How many moles of oxygen are required to burn 34.2 g of methyl alcohol?
2CH3OH + 3O2 --%26gt; 2CO2 + 4H2O


molar ratio O2 to CH3OH = 3:2


34.2g / 32.04 g/mol = 1.067 mol CH3OH


1.067 mol x (3/2) = 1.60 mol O2
Reply:Step 0: 2CH3OH (l) + 3O2 (g) --%26gt; 2CO2 (g) + 4H2O (l)





34.2g of methyl alcohol is 1.07 moles of methyl alcohol (I got that by dividing the mass by the GFM of CH3OH which is 32.05).





By looking at the balanced equation, every 2 moles of CH3OH takes 3 moles of O2 (g). So, if there are 1.07 mol CH3OH, there would be (1.07 x 3/2)mol O2 (g), which is 1.60mol O2(g).





Answer: 1.60mol or 51.2g of O2.





Here's the stoichiometry (contains steps 1-3 in brackets):


[34.2g * 1mol / 32.05g] * [3mol / 2mol] * [32.00g / 1mol]


= 51.2g of O2.

ivy