<p>ex.4. Sulfur trioxide, SO3, is produced in enormous quantities each year for use in the synthesis of sulfuric acid.</p>
<p>S (s) + O2 (g) —> SO2 (g)
2SO2 (g) + O2 (g) —> 2SO3 (g)</p>
<p>What volume of O2 (g) at 350deg C and a pressure of 5.25 atm is needed to completely convert 5.00 g of sulfur to sulfur trioxide?</p>
<p>Solution:</p>
<p>5.00g S x (1mol S)/(32.07g S) x (1 mol O2/1 mol S) = 0.156 mol O2</p>
<p>PV= nRT
V = nRT/P = (0.156 mol O2)(0.08206)(623K)/5.25atm = 1.52L O2</p>
<hr>
<p>Do I need to use the second equation? What about the “sulfur trioxide” at the end of the question? Are those “excess” information or am I missing something?</p>
<p>It’s asking how much oxygen gas (O2) is needed to convert S to SO3. In other words, there are 3 mols O per mol S, or 3/2 mol O2 per mol S. Using 3/2 you could do it in one step.</p>
<p>However to make it easier, you can take do it in two steps using the two equations given.</p>
<p>Look at equation 1: S + O2 -> SO2. If you have 5g sulfur, you have .15625 mols. The mole ratio for S:O2 is 1 (one mole of O2 per mole of S), so in other words you need .15625 mols O2 as well. I think you got that part, but you only showed how much O2 was needed to convert 5g of sulfur to SO2, and you didn’t do the second part.</p>
<p>Now look at equation 2: 2SO2 + O2 -> 2SO3. Here, in order to react, for every 2 mols of sulfur (2SO2 + O2) you need one mol O2. Thus the ratio is 1/2. Since you have .15625 mols S (remember from previous equation), divide that by 2 to get how many mols of O2 you need. It comes out to .078125 mols O2.</p>
<p>Add together the mols of O2 needed for equation 1 and the mols O2 needed for equation 2 and you get a total of 0.234375 mols O2.</p>
<p>Now put that into the gas law equation. (5.25atm)V=(.234375mol O2)(.0821)(623K) and solve for V.</p>
<p>The answer should be 2.283L of O2.</p>
<p>Wow. Thank you very much. Your explanation was really clear to me. I understand my mistake now.</p>