At one time, a common means of forming small quantities of oxygen gas in the laboratory was to heat KClO3. \n2 KClO3(s)2 KCl(s) + 3 O2(g)     H = -89.4 kJ\n\n(a) For this reaction, calculate H for the formation of 5.22 mol of O2.\n  kJ\n(b) Calculate H for the formation of 241.5 g of KCl.\n  kJ
119746
The reaction should be written as follows,\n2 KClO3(s) --> 2 KCl(s) + 3 O2(g) H = -89.4 kJ\n\nYou either did not include the arrow or it is not displaying properly.\n\nPotassium Chlorate decomposes with heat to form Potassium Chloride and Oxygen gas, no where in this reaction is elemental Potassium involved.\n\nThe standard enthalpy of formation of any pure element in its natural state is zero. Oxygen is a diatomic molecule and thus, Oxygen gas (O2) has a enthalpy of formation of 0 kJ in this reaction, no matter how much of it you have, it is still zero.\n\nSince the Potassium Chloride is the only product left that has any non-zero enthalpy of formation, we know that the over all change in enthalpy is equal to the enthalpy of formation of KCl * the number of moles produced.\nIf the over all change in enthalpy is -89.4 kJ and 2 moles of KCl are produced, then the enthapy of formation of the KCl must be 44.7 kJ/mole less than the enthalpy of formation of the Potassium Chlorate.
86930
Hess's Law says we can use all 3 reactions to get get our product, but H will be the same regardless of the way.\n\n2S(s) + 3O2(g)-> 2SO3(g) ΔH= -790kJ\n2SO2(g)-> 2S(s) + 2O2(g) ΔH= 594kJ \n\nUsing these 2 reactions, the final reaction will be:\n\n2SO2(g) + O2(g)-> 2SO3(g) ΔH=594-790=-196kJ\n\nSo ΔH=-196kJ
536570
first, write down the reaction:\n 2C3H8 + 7O2 -> 6CO2 + 8H2O\n\nthen, convert gram of C3H8 to mol -> 160/44 = 3.64 mol.\nthird, find the number of mol needed to burn C3H8 by looking at the coefficient of the O2 and C3H8 in the reaction.\nn of O2 = 7/2 * 3.64 = 12.74 mol.\n\nafter getting mol of O2, we can compute the volume of O2 using:\nPV = nRT. remember that the temperature is in Kelvin not in Celcius. the pressure must also be in atm not in Pa.
495739
2KOH=>K(2)O+H(2)O\n\nFeO+C=>CO+Fe\n\nCa+2AlCl=>CaCl(2)+2Al\n\nP(2)O(5)+3H(2)O=>2H(3)PO(4)\n\n2KI+Pb(NO3)2=>2KNO(3)+PbI(2)
650014
Problem \ntake D as Delta(change)\nEstimate the change in enthalpy, DH, for the following reaction:\n\nH2(g) + Cl2(g) --> 2 HCl(g) \n\nSolution \n\nTo work this problem, think of the reaction in terms of simple steps:\n\n\nFirst, the reactant molecules, H2 and Cl2, break down into their atoms\nH2(g) --> 2 H(g)\nCl2(g) --> 2 Cl(g)\n\n\nNext, these atoms combine to form HCl molecules\n2 H(g) + 2 Cl(g) --> 2 HCl(g) \n\nIn the first step, the H-H and Cl-Cl bonds are broken. In both cases, one mole of bonds is broken. When we look up the single bond energies for the H-H and Cl-Cl bonds, we find them to be +436 kJ/mol and + 243 kJ/mol, therefore for the first step of the reaction:\n\nDH1 = +(436 kJ + 243 kJ) = +679 kJ\n\nBond breaking requires energy, so we expect the value for DH to be positive for this step.\n\nIn the second step of the reaction, two moles are H-Cl bonds are formed. Bond breaking liberates energy, so we expect the DH for this portion of the reaction to have a negative value. Using the table, the single bond energy for one mole of H-Cl bonds is found to be 431 kJ:\n\nDH2 = -2(431 kJ) = -862 kJ\n\nBy applying Hess's Law, DH = DH1 + DH2\n\nDH = +679 kJ - 862 kJ\n\nDH = -183 kJ\n\n\nAnswer\n\nDH = -183 kJ
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This is a stoichiometric problem.  \n\nFirst, find the the amount in grams of one mole of Fe2O3.\n\nFor the first problem, convert the 3.02g Fe2O3 to moles.\n\nThen set up dimensional analysis.  For every 2 mol of Fe2O3, you need 2 mol C.  So you need an equal amount of moles for both.\n\nTake the amount of moles you have of Fe2O3 and convert them to grams Carbon.  \n\nFor the second problem, you should do the exact same thing except you dont need to convert to moles in the beginning because it's already in moles!\n\ni hope this is helpful!!!
173450
Write the equation for K.  You know K, [NO2}, so solve for the only variable you don't know [N2O4].  By the way, these are gasses so it is not a solubility product.  That involves a solid.
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When you have weak electrolytes, (equillibria in general) you should make the following table.\n\n1. write the reaction\n\nHNO2 <=> H+ + NO2-\n\n2.Put the initial amounts (expressed as concentration) of each molecule. In this case we have only the concentration of HNO2, so\n\n.. .. .. .. .. .. HNO2 <=> H+ + NO2-\nInitial .. .. .. .. 0.5\n\n3.Assume that an unknown amount "x" of the reactants react.\nThen write also how much of each product you expect to be produced according to the stoichiometry. In this case 1 mole of HNO2 gives one mole of H+  and NO2-, so x mole/L HNO2 will  give x mole/L of each and our table becomes\n\n.. .. .. .. .. .. HNO2 <=> H+ + NO2-\nInitial .. .. .. .. 0.5\nDissociate .. ..x\nProduce .. .. .. .. .. .. .. ..x .. .. .x\n\n4. Ka is an expression for the concentrations at equilibrium. So let's see what are those. For the reactants you will have initial-what reacted. We have only HNO2 with 0.5-x. For the products it will be the initial+what was produced. We didn't have any intial so it is only what was produced (x for both of the products). Thus the table in its final form is \n\n.. .. .. .. .. .. HNO2 <=> H+ + NO2-\nInitial .. .. .. .. 0.5\nDissociate .. ..x\nProduce .. .. .. .. .. .. .. ..x .. .. .x\nAt Equil. .. .. 0.5-x .. .. .. x .. .. .x\n\nKa= [H+][NO2-]/[HNO2] = x^2/(0.5-x)\nNow you either solve the quadratic equation and find x, or we assume that 0.5 >>x and practically 0.5-x=0.5.\nThen Ka=x^2/0.5=4*10^-14 =>\nx=squareroot(2*10^-14)= 1.4*10^-7. It is much smaller than 0.5 so our assumption is valid (otherwise we would have to solve the quadratic and get an exact solution).\n\nFor a simple level of chemistry we would now say\npH=-log[H+]= -logx =-log(1.4*10^-7) = 6.85\n\nHowever for more advanced and accurate expectations we see that the H+ coming from HNO2 is found to be in the same order of magnitude as that coming from the self dissociation of water (10^-7 for distilled water). Therefore in this case we cannot ignore the self dissociation of water (as we usually do) for calculating the pH. \nThere are 2 approaches for this case, here is one of them.\nRemember that we didn't put an initial amount for H+ since in most cases we ignore the self dissociation of water. Now we will consider that we had pure water, so the initial concentration of H+ is 10^-7 and we want to see how much the HNO2 will dissociate under these conditions so our table becomes\n\n.. .. .. .. .. .. HNO2 <=> H+ + NO2-\nInitial .. .. .. .. 0.5 .. .. .. 10^-7\nDissociate .. ..x\nProduce .. .. .. .. .. .. .. ..x .. .. .x\nAt Equil. .. .. 0.5-x .. 10^-7+x. .x\n\nand Ka=(10^-7+x)x/(0.5-x)\nNow we have to solve the quadratic so we can't do any simplification\n\n0.5Ka-Kax=10^-7x +x^2 =>\nx^2+ (10^-7+Ka)x-0.5Ka=0\n\nThe only acceptable solution is the positive one\nx=0.5(- (10^-7+Ka)+ squareroot (D))\nwhere D=(10^-7+Ka)^2-4*(-0.5Ka)=\n=(10^-7+4*10^-14)^2-4*(-0.5*4*10^-14)\n=9*10^-14\n\nand x=0.5*(-((10^-7))+4*10^-14) +3*10^-7) =10^-7\n\nso now pH=-log(10^-7+x) =-log(2*10^-7) = 6.70
430104
im pretty sure but  not positive\nC3H8 + 5 O2  -->  4 H2O + 3 CO2\n\nso 5 mol O2 to 1mol C3H8
160066
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
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Thw formular to use here is Q = mcT where Q = quantity of heat, m = mass, c = specific heat capacity, and T is the change in the temperature, thus here Q = 2.0g*(30-25)*0.90 = 9J. Notice that I sed Joules as the unit.\n\n2.For this one, a law which i cant remember the name right now is nedeed. Here we have C(s)+O2(g)=CO2 Enthalphy -390KJ\nand also Mn(s)+O2(g)=MnO2(s) Enthalphy = -520KJ\nFor this law to be appplied, we have to write the second equation the other way to become MnO2= Mn+O2 Enthalphy = 520KJ. Notice that i changed the sign when i reversed the equation. Then O2 could be canceled out on the left side of the first equation and on the right side of the second equation. This was we are adding. Finally adding, we get MnO2(s)+C(s) = Mn + CO2 Enthalphy = -390KJ+520KJ = 130KJ.\nand thats the correct answer
181900
This formula might help\n\nchange in S=(m1)(c1)(ln(Tf/Tc))+(m2)(c2)(ln(Tf/Tc))\nm1 & m2=substance mass in this case oxygyn gass (O2)=15.999(2)\nc1 & c2 are spcific heats\nTc =cold temp\nTh = Hot Temp\nTf=((m1)(c1)(Tc)+(m2)(c2)(Th))/((m1)(c1)+(m2)(c2))\n\nI'm assuming m^3/kg is the pressure (Volume/Mass) ???\n\nhope this helps!!
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The KC value can be answered by knowing the equilibrium concentrations of NH3,N2 and H2.\nSince 2 mole of NH3 was taken initially and 1 mole of NH3 remains unreacted so moles of NH3 reacted :1.\nSince 1 mole of NH3 has reacted it means that 0.5 mole of N2 and 1.5 moles of Hydrogen has been formed in the reaction at equilibrium.\n\n              we have \n                            kc=[N2][H2]^3/[NH3]^2;\n                            [N2]=0.5;\n                            [H2]=1.5;\n                            [NH3]=1;\n                therefore\n                            kc=(0.5)*(1.5)^3/(1)^2;\n                 which gives Kc=1.6875\n                clarification to your doubt:\n                since the reaction taking place is:\n                    2NH3 <----->N2 + 3H2. \nthis means    NH3 <----->0.5N2 + 1.5H2. (divide the equation by 2 \n to get moles of N2& H2 produced per mole of NH3)\n   As 1 mole of NH3 reacted which means that 0.5 mole of N2 and   1.5 mole of NH3 is formed at equilibrium.Just substitute in Kc and find the  value as 1.6875.I think that answers your doubt.
537979
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
227207
You need to take more care in writing equations when you post your questions. \n\nPart a) is difficult to decipher - you will need to re post it.\n\nPart b) is presumably:\n\nCH3COOH(l) + 2O2(g) --> 2CO2(g) + 2H2O(l)\n\nPart c) is presumably:\n\nFe2O3(s) + 3CO(g) --> 2Fe(s) + 3CO2(g)\n\nI would next use heats of formation to calculate the enthalpy change in the above reactions using the formula:\n\ndelta Ho = sum(heats of formation of products) - sum (heats of formation of reactants).\n\nIf you have a data book or data sheet, these values can be found there (Note: heats of formation of elements are zero.) \n\nIf you want web sites that might help you, try googling "enthalpy change problems" +tutorial.\n\nI hope this helps you.
553353
You need to find the heats of formation of each species (will come from a standard table) and use this equation to get the heat of reaction:\n\n(delta)H = sum of heat of formation of products - sum of heat of formation of reactants\n\nEach heat of formation will be multiplied by its stoichiometric ratio, and the physical state will be important (especially that of water, as the heat of formation of steam is higher than that of liquid water).  You will not find a heat of formation for oxygen gas, because by definition there is no heat of formation for an element.
474493
To answer this, you can use the ideal gas law, V = nRT/P\n\nYou know R , T  and P, andso you need to find n.\n\nTo find n , note that 2 moles of KClO3 produces 3 moles of O2. Given the mass of KClO3, you kind find the moles of this compound by dividing the mass by the molar mass. Assuming that there is 100% conversion, you can use the ration above to find the moles of O2.\n\nLastly, I'm not sure, but I think you need to use the gage pressure, which is the barometric pressure - the vapor pressure.