<p>4C2H5 + 13O2 >> 10H2O + 8CO2</p>
<p>calculate the mass in grams of O2 required for the complete combusiton of the sample of the hydrocarbon.</p>
<p>plzzzzzzzzzz help</p>
<p>I will return the favor in some way (chances, help w/ something else, etc)</p>
<p>how much hydrocarbon do u have...thats not enuff info</p>
<p>the problem doesn't state how much hydrocarbon...that's why I'm having so much trouble w/ this problem.... </p>
<p>should I take the molar mass of the hydrocarbon in the equation and use that as the amt. of hydrocarbon?</p>
<p>someone, plz. help...</p>
<p>I assumed 1 mol of hydrocarbon... multiply by mole ratio of O2 to the hydrocarbon, and then converted mol of O2 to g of O2. Just simple dimensional analysis.</p>
<p>Got 104 g as my answer.</p>
<p>can any confirm this answer/method?</p>
<p>Yes, as stated previously it depends on the amount of the hydrocarbon. If it is 1 mole, then calculate the equivalent O2 needed. Since the ratio of hydrocarbon to oxygen is 4/13 then the ratio dictates that for 1 mole of Hydrocarbon you need 3.25 moles of 02. 3.25 8 32 = 104 grams</p>
<p>A general equation you can plug in any mole of Hydrocarbon would be
(4/13)(n)(32g) equals grams of O2 for any n number of moles of hydrocarbon.</p>