<p>Post your questions here so that others may contribute. Perhaps this thread could make a useful tool for studying for the AP exam in May.</p>
<p>Here’s my question:</p>
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<p>My friends and I produced answers ranging between 40 kg and 90 kg of carbon. What did you get?</p>
<p>I’ll try it…</p>
<p>255 mi / (20.5 mi/gal) = 12.439 gal * 3785 = 47081 ml C8H18 * (1 g/ 0.69 ml) = 68233 g C8H18</p>
<p>68233 g C8H18 * ( 1 mol C8H18/ 114.26 g C8H18) * (16 mol CO2/ 2 mol C8H18) * (44.01 g CO2/ 1 mol) = 195920 g CO2</p>
<p>195920/1000 = 195.92 kg CO2</p>
<p>I had to look up gallons to ml though, I don’t know if you’re given that info.</p>
<p>I think I gave the incorrect value of miles traveled. It should be 225 mi.</p>
<p>We were allowed a unit conversions reference sheet.</p>
<p>I’ve determined the correct answer:</p>
<p>First, find the amount of fuel that was burned by dividing 225 mi by the gas mileage 20.5 mi/gal and find that 10.9756 gal of octane was burned.</p>
<p>Second, determine how much octane is present per liter of fuel by multiplying the density 0.69 g/mL by 1000 mL to get 690 g per 1 L fuel.</p>
<p>Third, complete the stoichiometry as follows:</p>
<p><a href=“10.9756%20gal%20octane”>quote</a> x (3.78541 L octane / 1 gal octane) x (690 g octane / 1 L octane) x (1 mol octane / 114.23 g octane) x (16 mol carbon dioxide / 2 mol octane) x (44.01 g carbon dioxide / 1 mol carbon dioxide) x (1 kg carbon dioxide / 1000 g carbon dioxide)
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<p>= 88.3591 kg of carbon dioxide</p>
<p>I used 1 g/.69 ml instead of .69 g/1 ml… derp.</p>