Suppose I burn 15 grams of butane in 500 grams of oxygen gas. One of the reactants will be completely used up and one will be left over.
1) Which reactant will be used up?
2) Which reactant will be left over?
3) How many grams of this reactant will be left over?
(Hint: Write a balance chemical equation first! Then think it through from there.)
c4h10 is butane
ReplyDeleteIt is with a capital C and a capital H.
ReplyDeleteSince this formula is being burned, this must be a combustion reaction. The products are CO2 and H2O. The formula would have to be C4H10 + O2 (arrow) CO2 + H2O. May someone else please balance this?
ReplyDeleteI think that the balanced formula would be 2C4H10+13O2 --> 8CO2 + 10 H20 because to balance you would first get C4H10+6.5O2-->4CO2 + 5 H20 but because there is a decimal you have to multiply the entire equation by 2.
ReplyDeleteFinding the moles of each substance will help. The molar mass of C4H10 is 58.14 g/mol, so the moles of butane in 15g of substance = 0.26. The molar mass of O2 is 32.00 g/mol, so the moles of oxygen in 500 g of substance = 15.63.
ReplyDeleteDon't we need to know the molar mass of the reactants and products in this equation? If so, 2C4H10 has a molar mass of 116.28 g. 13O2 is 416.00 g. 4CO2 is 176.04 g and 5H2O is 90.1 g. I hope this helps!
ReplyDeleteTo elaborate more onto Jordan's comment, butane would be an example of a hydrocarbon, and therefore, carbon dioxide and water would have to be the products of this combustion reaction. Also, the balanced formula looks good Juliette! Maybe butane being a hydrocarbon is an important factor in this problem...
ReplyDeleteBased off the balanced formula jordan compiled, it is agreed that it is a combustion reaction since there are carbon and hydrogen as reactants however what would be the reactant that is completely used up. Would this fact make it a combustion reaction?
ReplyDeleteTo figure out how many grams of oxygen gas might be left over, I used the above mentioned data for the amount of moles of butane in 15 grams of substance to be 0.26. Also, I used the molar mass of 13O2 to be 416.00 grams. Here is what I tried: 0.26 moles of C4H10 multiplied by 416.00 grams of oxygen/1 mole of C4H10. I got 105.56 grams of oxygen to be my answer. I hope this helpful!
ReplyDeleteYou are on the right track, Serene, but your calculations are missing a step. How could you figure out how many moles of oxygen would react with your 0.26 moles of butane?
DeleteGoing back to what Ambreen said, butane would be the reactant that is completely used up. It would still be a combustion reaction because the leftover oxygen would not be written as a product in the chemical equation (because it doesn't take part in the actual reaction) so the only two products would still be H2O and CO2.
ReplyDeleteNice, Colin, exactly right.
DeleteTo summarize thus far: we have concluded that butane is the reactant used up. Oxygen is the reactant left over. According to Serene´s calculation, there will be 105.56 grams of oxygen left over. What does everyone think about these results? To be honest, I am a bit confused on how we were able to distinguish which reactant was left over. Can anyone explain this?
ReplyDeleteExcellent question.
DeleteI agree with what everyone has been saying as far as this being a combustion reaction. Any reaction where the reactants are composed of carbon, hydrogen, and oxygen must be a combustion reaction (or at least I think so!), so the products would be H2O and CO2 like Colin said.
ReplyDeleteI agree with Colin, this is still a combustion reaction because the final products would still be water (H20) and carbon dioxide (CO2). I also got the same answer as Serene, there are 0.26 moles of C4H10 in 15 grams and I multiplied that by 416.00 g oxygen divided by 1 mole of C4H10. My final answer was also 105.56 g O.
ReplyDeleteSee my comment under Serene's post above.
DeleteI agree with all who said that this is a combustion reaction. I also agree with the equation Jordan came up with. I agree with Serene's calculation. Now I think we have to use the lesson we learned today. We have to use the mass given to us and use it to find the moles. Then we can use the lesson we learned today and see how much of the reactant we used and left over. If we find all of this, question 3 can be answered.
ReplyDeleteI agree with everyone so far! This is a combustion reaction. Also, the butane is used and the oxygen is left. I think that the O2 would have only one sig fig in the answer because we start off with 500g of it, which only has one sig fig so the proper moles of O2 should be 20 moles (I think...) I got the same answer as Serene, there are 0.26 moles of C4H10 in 15 grams. However, Georgia, I am a little confused on how you got your answer of 416.00 g oxygen and then went on to get the final answer. Could somebody please explain! Thanks so much.
ReplyDeleteApplying what we learned in class today to the balanced equation posted above, 13 moles of oxygen are needed for every 2 moles of butane, or 6.5 moles of oxygen for every mole of butane. Since we have 0.26 moles of butane, only (0.26*6.5=) 1.69 mol oxygen are needed for this reaction, which means (15.63-1.69=) 13.94 moles are left over. Converting this to grams will give 446.08g of oxygen left over.
ReplyDeleteNice, Valerie, you included the missing step.
DeleteCan someone please explain how to find the used up and left over reactants? Is there a different vocabulary word we used in class to describe this? Other than that I agree with the formula, the balanced formula, and the conversions used to find our answer 105.56 g O! I think we just need to clarify and do a small summary for people who don't quite understand the vocabulary.
ReplyDeleteThe leftover reactants are whatever doesn't get used up in the reaction. To burn the butane, we need 13 moles of oxygen for 2 moles of butane. But since we have only 15 grams of butane and 500 grams of oxygen, once we burn up all the butane there's still a bunch of left over oxygen that didn't get burned up. To find it, we subtract the starting amount (500 grams O2) by the amount we used to burn the butane. Hope this helps!
DeleteGreat point, Kevin. There is vocabulary that we haven't learned yet that goes with this whole problem. The reactant that gets used up is called the "limiting reactant". The reactant left over is called the "excess reactant". Brandon, you summarized the problem beautifully.
DeleteGood job, guys! I agree with everyone in that the equation is definitely a combustion reaction, and Juliette did a good job balancing it. . .I also get how we figured that butane is the reactant used up and oxygen is the reactant left over, I'm just a little confused over how Georgia and Serene solved for the grams of oxygen left over. Why would we multiply the 0.26 moles of C4H10 by 416.00 g oxygen?
ReplyDeleteOh, Valerie I missed your comment. Sorry! Never mind
DeleteThis comment has been removed by the author.
DeleteAh ha! Great analysis of where the logic breaks down, Lauren. And then you found the solution in Valerie's comment.
DeleteI see how combustion must be the reaction in this problem, and applying the lesson we learned today I agree with the conclusion that Serene made. I just want to elaborate that when converting to grams we used the molar mass of oxygen, which was 16.00 multiplied by the number of oxygen atoms.
ReplyDeleteI understand the steps that Serene took, and I also agree with the conclusion that she made. But just a general question... Are the products for a combustion reaction always H2O and CO2?
ReplyDeleteI believe that all combustion reactions with hydrocarbons (a compound made of only hydrogen and carbon) always has products of H2O and CO2, but only if it's a hydrocarbon and oxygen.
DeleteYes indeed, Brandon, and the same is true for hydrocarbons like glucose, C6H12O6, which contain oxygen in addition to the carbon and hydrogen. The products are always carbon dioxide and water.
DeleteI understand all of the calculations and the balanced equation. What I do not understand is how we know that it is a combustion reaction. Can someone please explain?
ReplyDeleteWe know that this is a combustion reaction because if the formula is being burned, then it must be a combustion reaction. Also, if you look back at the handout on the types of reactions, if carbon, hydrogen, and oxygen are all reactants, then it must be a combustion reaction (aka CHO reaction)
DeleteNice work, team! You are nearly there. Serene and Georgia made great strides in the calculation, but their logic is missing a step. Valerie's comment identifies the missing step. Can I get someone to summarize what the missing component in Serene and Georgia's logic was and how Valerie's logic addresses it?
ReplyDeleteValerie's logic addressed the problem because we had to find all of the water that was left. Like Brandon said, after we burned all of the butane, we still had a lot of O2 left. So, Valerie took the amount of oxygen that we were given in the problem, 500g O2 and found the moles of water that it had, which is 15.63 moles of O2. Then she subtracted that by 1.69 moles of O2 (because this is what is left after the butane is burned off.) So, she subtracted to get the amount of O2 that was left, turned it into grams and you get 44.08 g of O2 left over.
ReplyDelete