I don't get Gas Stoichiometry!!! Anyone know anything?
367911
For example, hydrogen burns in oxygen to form water vapor. \n\n2H2(g) + O2(g) -> H2O(g) \nHow much hydrogen must be burned to form 1.00 L of water vapor at 1.00 atm pressure and 3000C? \nTo work this problem, we need to figure out how many moles of water vapor we need to form. Use the ideal gas equation to compute this: \n\nPV = nRT \n1.00 atm * 1.00 L = n*0.0821 (L*atm/mol*K)* 473 K \nn = 0.0258 moles \nThus, we need to form 0.0258 moles of water. Now it's just a basic stoichiometry problem: for each mole of water formed, you need 2 moles of hydrogen gas, so 0.0258 moles water * 2 moles H2/1 mole water = 0.0516 moles of hydrogen \n\n=====================================================\n\nthis will help you a lot\n\nhttp://www.chem.hawaii.edu/asato161/Ch10NotesP2.pdf#search='Gas%20Stoichiometry%20PV%3DnRT'\n\nhttp://members.aol.com/profchm/gasstoic.html
963278
1) Determine how many moles of CH4 we have:\n175 g/ (12.01+4*1.008) =10.9 mol\n\n2) Write up to reaction to determine the amount of O2 in moles:\nCH4 + 2 O2 ->  CO2 + 2 H2O\nWe need 2*10.9 = 21.8 mol of O2\n\n3) The volume of 1 mol gas is 22.4 liter at STP. At the given conditions:\nV(1mol) = 22.4 * (290/273) = 23.79 liter\n\nTherefore 21.8 mol = 519 liter = 0.519 m^3
761497
1) The first one is wrong on the last person who answered, O2 weighs 32g/mol because there are 2 O's in it, 16 each.  So in 1.33x10^4 (13,300mol) you have 4.256x10^5g of O2 (425.6kg).\n\n2) Recognize that hydrogen, oxygen, and chlorine are diatomic gasses, so they always occur in pairs (H2, O2, Cl2).  Others like this are nitrogen (N2), fluorine (F2), bromine (liquid @ room temp. Br2) and Iodine (solid at room temp, I2).  So your equations would be:\n\n2H2+O2-->2H2O\nH2+Cl2-->2HCl\n\n3) PV=nRT and solve for n, number of moles.  Put everything in correct units, liters, atmospheres, moles, and kelvins, and R is a constant R=0.0821 L*atm/mol K.  Kelvin temp=Celcius temp +273\n\nSolving for n:\nn=PV/RT=(1.09atm)(0.942L)/(0.0821L*atm/mol*K)(287K)\nn=0.0436 mol\n\nSo, when we usually want to know a number of moles given a mass and having a molar mass we can look up, we say,\nn=mass/molar mass\n\nin this case, we need to find molar mass, so:\nmass/n=molar mass\n3.17g/0.0436mol=72.71g/mol
655398
Here's how to solve it (I am convinced that you should do your own homework, but hints are OK as long as you learn something):\n\nStart solving by finding the molecular mass of the compound.\nA. Volume occupied by one mol of gas is 22.4 L at standard temperature and pressure (STP: 760 mmHg and 273 K (0 C))\nTo find the volume of gas that weighs 2.57 g at STP, use formula: P1V1/T1=P2V2/T2, where R is 0.08206, P1=319 mmHg, V1=2 L, T1=298 K; P2=760 mmHg, T2=273 K and V2 is unknown (it will be somewhat less than 1 L).\nAfter you get this volume, see how many mols do you have in it: divide 22.4 by V2 you just got. Then multiply this number by 2.57 and get the molecular mass of your compound (M(X)).\n\n2. Multipy M(X) separately by 0.6481, 0.136 and 0.2159 to get number of atomic weight units that are allocated to C, H, and O respectively. \nThen divide first number by 12, 2nd by 1 and 3rd by 16 to see how many atoms it amounts to. The resulting integer numbers are X, Y, Z.\n\n3. CxHyOz is the empirical formula of your compound. Google it to get the possible molecular ones. This is a trick question, so there will be more than one but less than 8. All of them are alcohols or ethers.
194284
To solve this problem we need to think about limiting reactants.  This one is fairly simple because I already have moles and the ratio is 1:1:1.  If I only have 3 kmols of C, that is the most I can make, because going beyond that, I would be out of C.  Thus, I can only make 3kmols of CO2 with 2 kmols of O2 left unreacted.  These amounts are theoretically find for a combustion reaction.
878226
The method is very simple.  I can do the first and you can try the others.\n\n1.  First write the reaction and balance it.\n\n2HgO ---> 2Hg + O2.\n\n2. Convert grams to moles of whatever mass you have.  In this case you have O2 as 96 grams.  You convert to moles by dividing the mass by the molecular weight.  In this case, for O2 will be 32 g/mol\n\n96 grams / 32 g/mol = 3 mol.\n\n3. Now you look back to the equation.  In the equation for producing 1 mol O2 you need 2 moles of HgO.  If now have 3 moles O2, then you will need 2 x 3 moles of HgO, 6 moles.\n\n4. Now you need to convert the moles of HgO to grams.  To do this calculate the molecular weight of HgO.  It is 216.6 g/mol\n\n6 moles * 216.6 g/mol = 1299.6 g of HgO.
50447
Hey Dude! Do you know how many different types of gases there are!!!! Do you! I can't even list all the types of gases here! Jeez!
1040448
ask your chem prof or tutor ..it's hard to explain over the internet!
250561
i can help with part of it. i forgot a lot of my chemistry stuff. so ok  first thing is to find the number of mols of each reactant put in. which is  mass over molar mass. and then find the molar mass of 0.566g of H2. so you know how many moles of H2 are left over, so you should lknow how many moles reacted and you should know the number of mols of the other 2 substances too and you know the volume and you can figure out the masses so then   do that. \nand Kc is just the  conc of products over reactants. if youre in AP   i assume you know how to do that\nthen c   and    um partial pressures. i think it has something to do with mole ratios and   the total pressure. im not completely sure, but im sure you can figure it out. good job
609831
pv=nrt\nmeasure pressure, volume, and temperature to find the number of moles, n\n\nfind the mass of the gas\n\nmass of gas/ moles of gas= molecular mass (g/mol)
558084
Start with the ideal gas law: PV=nRT.  Since no gas enters or leaves the system, n is constant, and R is the universal gas constant, the product nR has the same value at the beginning and as at the end.  So, collect it to one side, and move the other factors to the other side.\n\nP1 * V1 / T1 = nR = P2 * V2 / T2\n\nSolve for T2\n\nT2 = T1 * (P2 * V2) / (P1 * V1)\n= 300e4 K (.0003e15 atm * .046e-15 cc)/(50e3 atm * .0004e-7 cc)\n=2.07 * 10^7 K
793636
shawty's NOT good at stoichiometry...has made a fundamental error in the final line so end result is rubbish...\nhow can you end up with MORE Cu than you started with when it's 2 moles : 2moles? \n\n(8/159)x 127 = 6.4 g
914456
okay...100 g*(mol/12.01g)=8.33 mol C\n\ntake the aforementioned 1mol *(-113kj/2mol)=-56.5 kJ released for 1 mol\n\nthen multiply that -56.5 by 8.33= -470.65 kJ released\nnote it is exothermic and thus negative
1006740
The equilibrium constant for this reaction is given by:\n\n[NO]^2/([N2]*[O2]) = 6.91*10^-4\n\nYou also know, from mass balance, that:\n\n[NO] + 2*[N2] = 2*0.75 mol/L \nand\n[NO] + 2*[O2] = 2*2.4 mol/L\n\n (i.e., the total concentration of nitrogen and oxygen atoms doesn't change in this system -- the number of nitrogen atoms at equilibrium after the gasses react must be the same as the number of atoms you started with)\n\nYou now have three equations in three unknowns ([NO], [O2], and [N2]).  These three unknowns are the concentrations of the gases at equilibrium.  \n\nTo solve these equations, we can rewrite the mass-balance equations as:\n\n[N2] = 0.75 - 0.5*[NO]\n[O2] = 2.4 - 0.5*[NO]\n\nsubstitute these into the equilbrium constant equation to eliminate [N2] and [O2]:\n\n[NO]^2/((0.75 - 0.5*[NO])*(2.4 - 0.5*[NO])) = 6.91*10^-4\n\n[NO]^2/{1.8-1.575*[NO]+0.25*[NO]^2} = 6.91*10^-4\n\n[NO]^2 = 0.0012438 -0.001088325*[NO] +0.00017275*[NO]^2\n\n0 = 0.0012438 - 0.001088325*[NO] - 0.99982725*[NO]^2\n\nUse the quadratic formula to solve for [NO].  In general, there will be two roots, but only one will be physically meaningful.\n\n[NO] = -0.03582 or 0.03473.  Obviously, only the positive root has a physical meaning, so at equilibrium [NO] = 0.03473 mol/L\n\nSubstitute this into the mass-balance equations to get the concentrations of [O2] and [N2]:\n\n[N2] = 0.75 - 0.5*0.03474 = 0.73263 mol/L\n[O2] = 2.4 -  0.5*0.03474 = 2.38263 mol/L\n\nYou can check these results by plugging them back into the equilibrium constant equation to make sure everything agrees:\n\n0.03473^2/[0.73263*2.38263] ?= 6.91*10^4\n\n6.91*10^-4 = 6.91*10^-4  Check!
929547
I find 16.4% \nHere we go:\n1) As air is a mixture of O2 and N2 two reactions are taking place:\n2 Mg + O2 -> 2 MgO\n3 Mg + N2 -> Mg3N2 \nThere are 4 parts of N2 for one part of O2, so the second reaction takes place four times as frequently as the first. Effectively we have\n14 Mg + O2 + 4 N2 -> 2 MgO + 4 Mg3N2.\n\n2) The molar mass of MgO is 40.3 gram. In MgO the magnesium makes up 24.3/(24.3+16.0) = 60.3 % of the mass.\nThe molar mass of Mg3N2 is 100.9 gram, the magnesium makes up 3*24.3/(3*24.3 + 2*14.0) =  72.2 % of the mass. \n\n3) Assume you have 1000 gram of reaction product. If it is pure MgO you will find 603 gram of Mg. But in the mixture you will have  \ntwo moles of Mg2N3 (201.8 g) for every mole of MgO (40.3 gram).\nIn 1000 gram of mixture, 833.5 gram of  Mg2N3 and 166.5 gram of MgO. This gives 72.2% * 833.5 + 60.3% * 166.5 = 702 gram of Mg.\n\n4) Finally: the error is (702-603)/603 = 16.4%
294763
Well, stoichiometry is not a method or a device. It is simply calculations of reactions in chemistry. It doesn't really have a role but in smelting, since smelting is not a chemical change. It is actually a physical change of state.
275919
2H20 --> 2H2 + O2\n\ntheoreticalyy spk it's exactly half the vol of H2, just look over the stoechiometrical balance above; but due to gas losses, it may be less