It’s been a while since I did chemistry, but I don’t get why you have O[sub]2[/sub] on both sides of the equation. That seems to imply you’re always starting with 2:3 mixture of CH[sub]4[/sub] and O[sub]2[/sub], which isn’t even true for this specific case.
The reaction is just CH[sub]4[/sub]+2O[sub]2[/sub]–>CO[sub]2[/sub]+2H[sub]2[/sub]O. Since you’ve determined that the CH[sub]4[/sub] is the limiting reagent, start there and figure out how much of the O[sub]2[/sub] is consumed.
Oh…OK…yeah I screwed up the equation. I thought you had to pull it out of mid air turns out they posted the equation and I didn’t see it. I still wonder why the hell they ask you for the amount of CH4 left. To get you to figure out which one is the limiting reagent?
Thanks, I think that might do it. I’ll be back if it doesn’t.
Though you should have been able to figure out the equation quite easily. You’ve been told you’re burning CH[sub]4[/sub]. The four H’s turn into H[sub]2[/sub]O (two of them) and the lone C turns to CO[sub]2[/sub], using up four O’s (two O[sub]2[/sub]'s).
Like I said, by throwing in extra (unused) reagents on both sides of the equation, you’ve introduced some erroneous assumptions about the amount of each reagent.
If methane is the limiting reagent, then there’s going to be none left burning it in an excess of oxygen. That’s one answer sorted out. For the rest, use the correct eqn provided by scr4 and all should become clear.
Step 1: Write and balance the equation. Hint–for complete combustion of hydrocarbons (like methane) your only products are water and carbon dioxide. You should simplify the equation such that no species is on both sides of the equation. Your equation simplifies to:
CH[sub]4[/sub] + 2O[sub]2[/sub] -----> CO[sub]2[/sub] + 2H[sub]2[/sub]O Step 2: Convert grams of reactants to moles. You did this correctly for methane. It should be 0.5 moles. You did this incorrectly for oxygen.
6 g O[sub]2[/sub] x (1 mol O[sub]2[/sub]/32 g O[sub]2[/sub]) = 0.1875 mol O[sub]2[/sub]
Step 3: Determine the limiting reactant. Here’s a “trick.” Divide the moles of each reactant by its own coefficient. Whichever reactant results in the smaller number is the limiting reactant! However, DON’T USE THAT NUMBER FOR ANY CALCULATIONS. Only use the number to determine the limiting reactant.
Therefore oxygen is the limiting reactant, not methane!
Step 4: Determine moles of products formed (using molar ratios) based on moles of the limiting reactant that you started with. Here use the fact that you started with 0.1875 mol O[sub]2[/sub]. Figure out how many moles of carbon dioxide and water are formed.
Step 5: Convert moles of products to grams.
Step 6: If asked, calculate moles (and grams) of excess non-limiting reactant.
I taught general chemistry for over five years. If you still have questions, feel free to ask.
In addition to the issues raised above, Cyberhwk, your calculations in the OP assume that the molecular weight of CO[sub]2[/sub] is ~42. You might want to re-check that.
That’s an excellent point. However, you would produce a varying amount of CO that would not be able to be determined.
If I were writing the above problem, I would add “assume complete combustion.” Having written that, I would almost certainly not make oxygen the limiting reactant, because I like my problems to have some basis in reality.
If you look at the OP, the problem stated the species that were involved, and CO was not mentioned, presumably to accomplish the same aim.
I just wanted to say, I’m sure your grader is very grateful that you, first of all, did a reality check and realized that getting 96 grams of products out of 14 grams of reactants meant that you did something wrong, and second of all, that having determined that something was wrong, you went to some effort to find out what it was. I know that whenever I’m grading, I’m always pulling out my hair and going “Dear God, why!?” whenever I see that a student has made a mistake that should have been caught with a simple reality check like that.
