<p>hi, im using my bro acc to ask a question,</p>
<p>ok.. i havent been paying attention in class so im a little slow.. but i have a test tommrow, i want to master fundamentals first, so is this right?</p>
<p>using CO + O2 --> CO2, if you started with 120 grams of Carbon monoxide and 100 grams of Oxygen gas, which reactant would run out first?? i did</p>
<p>120g CO ( 1 mol CO/28g CO) ( 1mol CO2/1mol CO) ( 40g CO2/1 mole CO2) = 171.4</p>
<p>100g O2 ( 1mol o2/32g o2) (1mol co2/1 mol o2) (40 g CO2/1mol O2) = 125 </p>
<p>So O2 is the limiting reactant, is this right?</p>
<p>my bad CO is limiting mis write</p>
<p>Yes right answer but - you have to balance the equation first, so your O2 number is wrong. You can leave it in moles - no reason to go to grams unless they wanted to know how many grams of product were formed.</p>
<p>2CO + O2 --> 2CO2
120 g CO ( 1 mole CO/28 g CO) (2 moles CO2/2 moles CO) -> 4.286 moles CO2 product</p>
<p>100 g O2 (1 mole O2/32 g O2)(2 moles CO2/1 mole O2) --> 6.25 moles CO2 product</p>
<p>O2 is limiting</p>
<p>there is an extremely easy method to determining limit reactant... without doing anything</p>
<p>Here's the answer to your other question:
balanced Eq: 2C2H6 + 7O2 --> 4CO2 + 6H2O</p>
<p>Q. With an excess of oxygen gas and 7 moles of ethane, C2H6, how many moles of each product would you expect to make?
7 moles of C2H6 * (4 moles CO2/2 moles C2H6) -> 14 moles CO2
7 moles of C2H6 *( 6 moles H2O/2 moles C2H6) -> 21 moles H2O</p>