How can I calculate the heat released in the following chemistry equation?
816363
4 Fe(s) + 3 O2(g) ----> 2 Fe2O3(s)\n\nFirst, you have to check if the chemical equation is balanced. In this case, it is. So, what it tells you is that for every 4 mole of iron that is reacted, 1652 kJ will be released to the surrounding. Now, since O2 is in excess, iron will then be your limiting reagent. If you have 3.34 mole of iron, you can use ratio to determine how much energy will be released.\n\nE = (-1652/4) *3.34 = -1379.42 kJ\n\nNote that the "-ve" sign indicates energy is released instead of being absorbed.
812576
From the reaction you had heat produced\n\nQ=m*Cp*ΔT\n\nfor water Cp= 1 calorie/(gram*Celsius)\n\nYou have a total of 50+50 =100mL and since d=1.00 g/mL you have 100 g\n\nso Q =100*1*(31.43-25.32) = 611 cal\n\nyou mixed mole HCl= C*V= 1*0.05= 0.05 with\nmole NaOH =C*V =1.1*0.05 =0.055 which means that, since the stoichiometry of the reaction is 1:1, HCl is the limiting reagent\n\n0.05 mole HCl give 611 cal\n1 mole HCl gives x\nx=611/0.05= 12220 cal/mole =12.22 Kcal/mole
823187
Q( thermal equilibrium/heat  )= Mass *specific heat capacity * (Temperature final - temperature initial) This is the formula.
863200
Use the equation Q=mc(delta)T.\n36.8 = .006*130*(Tf - 32)\nSolve the equation for final Temperature and you get 79 degrees Celsius
1142668
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
366921
Q = mc delta T\nHeat = mass times specific heat times change in temperature\nspecific heat for water is 4.186 kJ/kg C\n-41840 J (negative because absorbed) = .200 kg x 4.186 x 10^3 J/ kg C x (Tf - 20)\nTf = final Temp.\nsolving the equation gives you Tf = -30.0 degrees Celsius
896861
Remember your equation for specific heat capacity::\n\nHeat content = mass x spec heat capacity x delta T\n\nq = m*C*dT\n\n225 J = (15 g) * (0.80 J/gram-deg C)* dT \n\nTherefore\n\ndT (in C) = 225 J  /  (15 g)(0.8 J/g-C) = 18.75 deg C\n\nMind your units and make sure they cancel.  I did.
773086
w = -(P x deltaV). Delta V is final V minus inital V (which you get by using the PV=nRT equation at each temperature). so -(3 x -11.8) = +35.39.\n\ndelta E = q + w. So add the q you got and the w you get.\n\ndelta H = delta E + (P delta V). \n\nThere ya go. hope this helps
475093
see dear ,\nwhile calculating the energy changes we need to know the hidden heat { latent heat } that is absorbed during the change in state . for eg when we are calculating the heat absorbed or given out during a change in state , like ice to water , just first calculate the latent heat ie \nq=mL{ heat=mass *latent heat } now add this value to \nq=mct{heat=mass*specific heat*change in temp} keep in mind that specific heat is a constant value \nwater's specific heat is 4180j/kg/degree Celsius.
85318
Heat is equal to the change in enthalpy, dH, which is equal to the enthalpy of the products minus the enthalpy of the reactants.\n\ndH=H(products)-H(reactants)\n\n2B + 3F(2) --> 2BF(3)\n\nGet the heat of formation values for each specimen from a chemical thermodynamics table. I don't have a table in front of me so I'm just going to make up values so you'll know how to solve the problem. Let H=100 for B, H=150 for F(2), and H=400 for BF(3).\n\ndH= 2(400)-[2(100)+3(150)]\ndH=800-650=150 kj/mol\n\nI do know that elements alone have a heat of formation of 0, but I don't remember if the element has to be in a certain state or not.
870267
.4084c(80)+.4084(heat of vaporization)\nc is specific heat of water\nthat will give you J/KG C   si units
963248
The heat given off by the burning reaction (delta E) is equal to the heat gained by the calorimeter.  (m * change in temp)\n\nYou can calculate the heat released by burning.\n.553 g with a molecular wt. of 88g/mole \n(.553 grams) / (88 (grams / mole)) = 0.00628409091 moles\n(975 (kJ / mole)) * 0.00628409091 moles = 6 126.98864 joules\n\nYou can calculate the heat absorbed by the water (3000 g * .356)\n3 000 * .356 = 1 068 joules\n\nyou can calculate the difference, it must have heated up the apparatus.\n6 127 - 1 068 = 5 059 joules \n\n5059 joules = x grams * .356\n5059 joules / .356 = 14210.67 grams of water\n\nThe metal apparatus was had a heat capacity equivalent to 14210.67 grams of water.
883220
H2+O2-->H2O\nNaOH+HCl-->NaCl+H2O\nKOH-->(on heating,deomposition)K2O+H2O\ni guess this is what u want\nand i f u didn't get it i am sorry......
737811
You use the formal \nH= mcdeltat\n\nmanipulate it to be...\n\nc (specific heat capactiy) = H /(mdelta t)\n\nChange temperature to Kelvin,....4degrees = 277Kelvin\n\nC=221/(55X277)\n\n\nC= 221/15235\n\nC=0.0145 (in JOULES)
800509
Here a total guess\n\nThe mole weight of C6H12 = 6*12 + 12*1 = ~84 grams\n3.57 grams / 84 grams = how many moles = M = 0.0425\n\nThe temperature rise = 31.3 - 29.1 = 2.2 degrees C = T\nThe heat capacity  = 6.210 KJ /degree = H\nSo the Energy released = H * T = 6.210 * 2.2 = 13.662 Kjoules\n\nSo the heat of reaction = H * T / M\nergo 13.662 / 0.0425 = 321.45882 KJ/Mole
184409
1.75mol  /  2mol of C6H6 times specfic heat ( -98.0kJ)\n\n=  .875 * -98.0kJ = -85.75kJ
533621
(delta)H(rxn) = sum of heat of formation of products - sum of heat of formation of reactants.\n\na.  Hf Br2(g) = 30.91 kJ/mol, Hf Br2(l) = 0, so (delta)H = -30.91 kJ/mol\n\nb.  Hf CaCO3(s) = -1207 kJ/mol, Hf CaO(s) = -635.5 kJ/mol, Hf CO2(g) = -393.5 kJ/mol, so (delta)H = -635.5 - 393.5 + 1207 = +178 kJ/mol\n\nc.  Hf NO(g) = 90.25 kJ/mol, Hf O2(g) = 0, Hf NO2(g) = 33.2 kJ/mol, so (delta)H = 2 * 33.2 - 2 * 90.25 = -114.1 kJ/mol