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Gas stoichiometry problem?

Gas stoichiometry problem? Topic: Problem solving with c by j
June 20, 2019 / By Ginnie
Question: The problem reads "2.34g C2H5OH are mixed with 5.6L oxygen at 78 degrees F and 900 torr. What mass CO2 can be made?" An unbalanced equation is given, too: C2H5OH + O2 ---> CO2 + H2O I've balanced it to what I think is correct. I have a 2 in front of the O2 and the CO2. I've made the necessary temperature and pressure conversions, too. I'm just not sure where to go from here to get to the answer. I know I have to use the PV = nRT equation at some point, I'm just not sure how to use it correctly in this problem. So, any help with it would be appreciated. Thanks a lot.
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Best Answers: Gas stoichiometry problem?

Denise Denise | 5 days ago
Do you have the answer to the question? When I balanced the equation I got C2H5OH + 4O2 ---> 2CO2 + 3H20... let's see, got 2 C's on both sides, 6 H's, 5 O's. yup. Seems balanced. Then, my oh my what silly units you use. I had to look up conversions for everything to get into normal units. Do you have to do that too? ok so let's see here... first do equivalents, 1 mol C2H5OH <=> 4 mol O2 <=> 2 mol CO2. Then. I've got 2.34g C2H5OH. First I think, how do I find the mass of CO2? mass = m... hmmm, m = M n is the equation I can think of. So that means I need n. The molar number for CO2. I know it's gonna be 2 times as much as C2H5OH, or half as much as O2, from the equivalents... So lets find out how many mol of C2H5OH we get out of 2.34grams of the stuff. n = 2.34g / (24 + 6 + 16)g/mol = 0,05 mol C2H5OH there are 0.05 mol of this stuff in 2.34 grams. So theoretically, we can find the answer to how much CO2 we need right now, BUT we gotta see which one is the limiting reagent. This, or oxygen. If this one runs out first, then we use this n-value to calculate CO2. If the other runs out first, then we use that n-value to calculate CO2, just using the same m = M n formula. Do you know which is the limiting reagent? To find this, we would take 0.05 mol and multiply that by 4 to get 0.2. We would need at least 0.2 mol of O2. If we get more than 0.2 mol of O2 then we know c2h5oh runs out first. If we get less, then we know that the O2 runs out first. So this is where we have to do all the pV=nRT garbage, just to test. So what are we after? the n. So solve for n. n = pV/RT p = 900 torr. which then equals 1.19x10^5 Pa (I think, according to some conversion online), V = 5.6 L = 5.6x10^-3 m^3, R is constant no matter what country you're in thank god, and that's 8.31 J/mol K, and T. Farenheit? What is the deal with that anyway? T = 298.56 K so plug it all in, blah blah blah and we get about 0.269 mol. n (O2) = 0.269 mol. That means we have more than 0.2 and therefore, we can go with the first (and easier) calculation of n to get CO2. So now we get m for CO2 m = M n (still) n (CO2) = n (C2H5OH) * 2 (from the equilibrium, from every 1 c2h5oh, we get 2 co2) so blah bah, m = 44g/mol * 0.05 mol * 2. Moles cancel out, we get grams. 4.4 grams. bah. I hope that's right. Otherwise I'm useless.
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Denise Originally Answered: Can anyone help me with this webassign chemistry problem on stoichiometry?
Liz From the balance equation you can see that 4 moles of NH3 will react with 5 moles of O2. But we don't know if the amounts of each of these substances is given in the correct molar amounts, so first convert the number of grams of each substance given into moles moles NH3 = l.88g over l7grams/mole = .11 moles ammonia moles O2 = 2.29 g over 32 grams per mole = .072 moles oxygen We now need to find out if these given number of moles of NH3 and O2 are in the correct ratio, or if not, which one will we run out of first. This one which runs out first is called the LIMITING REACTANT. because when you use it all up, the reaction stops. so we know we need 4 moles NH3 to 5 moles O2. we are given .11 moles NH3 and .072 moles of O2. Lets divide the smaller number of moles, .072 for O2 into the larger # moles of NH3 and see what kind of ratio we are given to work with This comes out l.53, so we have l.53 moles of NH3 for each mole of O2. This is the tricky part. We need a ratio from the balanced equation of 4 moles NH3 to 5 moles O2 we have l.5 moles NH3 for each l mole O2 After you look at the ideal ratio from the balanced equation and compare that with the ratio you are given you can see that you are short on oxygen, and oxygen will be your limiting reactant and will control how much product you can make. You can forget about the NH3 because you have more than enough to react will all the available O2. So use only the amount of O2 given in your calculations. Use the balanced equation above as a math platform for your calculations. Over the 5 O2 place 2.29 grams and under the 5 O2 place the total atomic weights of l0 oxygen atoms which will be 160 Now over the 4NO place X grams and under the 4NO place the sum of all the atomics weights in 4NO which add up to 120 Now cross multiply and solve for X grams of NO X = 1.72 gms NO Now over the 6 H2O place X grams and under the 6 H20 place the sum of all atomic wts in 6 H20 which is 108 Now cross multiply again between the oxygen numbers and the numbers above and below the H2O I get 1.54 grams water
Denise Originally Answered: Can anyone help me with this webassign chemistry problem on stoichiometry?
4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g) ..68.0 g......160 g...,.....120 g........108 g ? g H2O = 1.88 g NH3 x 108 g H2O / 68.0 g NH3 = 2.99 g H2O ? g H2O = 2.29 g O2 x 108 g H2O / 160 g O2 = 2.01 g H2O So O2 is the limiting reagent. Mass H2O formed is 2.01 g ? g NO = 2.29 g O2 x 120 g NO/160 g H2O = 1.72 g NO
Denise Originally Answered: Can anyone help me with this webassign chemistry problem on stoichiometry?
What`s the limiting reactant? You need ti know this as further calculations are based upon it. NH3: 1,88 g / 17.0 g/mol = 0.11 moles. O2: 2.29 g / 32.0 = 0.0716 moles. 0.11 moles NH3 would react with 5/4 * 0.11 = 0.1375 moles O2 which are not available. 0.0716 moles O2 would react with 4/5 * 0.0716 = 0.05725 moles NH3 which are availabke. O2 is the limiting reactant. 4/5 * 0.0716 moles = 0.0895 moles. 0.0573 * 30 = 1.72 g NO. 6/5 * 0.0716 moles = 0.0859 moles. 0.0859 * 18 = 1.55 g H2O.

Callie Callie
first get the molar mass of all 3. for each 2KClO3's you decompose, you get 3 O2's. Then i think you're taking an equation. PV=nRT rigidity must be converted to atm. 1520/760 gets that.(i've got not got a calculator on me) quantity often is the variable. R is .08206 each and every time T is 37+273, that's 310 i think of, featuring you with kelvins. n= the moles artwork the equation for V convert it to L via multiplying via one thousand lol, i did all that from reminiscence, so in case you think of i'm somewhat off or somebody has a contradicting answer, verify it.
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Callie Originally Answered: Omg.will u help me with my 8th graders math problem? It has algebra, fractions and is a story problem?
1miles / 8 minutes ( which is the speed of the person) * 1 minute = 1/8 this is how far you run in 1 minute 1 miles / 8 minutes (speed) * 11 minutes = 11/8
Callie Originally Answered: Omg.will u help me with my 8th graders math problem? It has algebra, fractions and is a story problem?
Note that you're asked to write an equation, not just say "1/8". Your speed is 1 mile per 8 minutes, or 1/8 miles-per-minute. Speed is also distance divide by time, which means distance is speed times time (in other words, if you multiply how fast you're going by how much time has passed, you'll get how far you've gone). If "m" is the number of minutes you've run, and "d" is the distance, then d = (1/8)*m. Another way to write this is d = m/8. Now that you have an equation, you put in any value for "m" and get back "d", the distance covered in m minutes. So for m=11, d=11/8, or 1 3/8 (one and three-eigths of a mile). Some people might look at this and say skipping the formula and just "dividing by 8" is quicker. But in science and engineering, sometimes you have lots of different variables which means you need an equation instead of trying to do everything in your head. So this is an important concept to understand. On a side note, if you go around saying you "loathe math", chances are your 8th grader might get the impression that somehow it's OK to be "bad at math", and thus isn't going to want to excel at math either. Something to think about.

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