The reaction below is at equilibrium. Use LeChatelier’s principle to predict the effect of removing nitrogen gas to the equilibrium reaction mixture.\nN2(g) + 3H2(g) = 2NH3(g)
198771
Based on LeChatelier's principle, removing Nitrogen (N2) from the above equation would cause the equilibrium to shift towards the left.
504003
Part A:\nThe answer is 12 L H2.  Here is the solution:\n(4 L N2)(1 mol N2 / 22.4 L N2)(3 mol H2 / 1 mol N2)(22.4 L H2 / 1 mol H2)\nso (4*1*3*22.4) / (22.4*1*1) = 12 L H2\n**note that in in ideal gas there are 22.4 liters/mol**\n\nPart B:\nThe answer is 8 L NH3.  Here is the solution:\n(4 L N2)(1 mol N2 / 22.4 L N2)(2 mol NH2 / 1 mol N2)(22.4 L NH3 / 1 mol NH3)\nso (4*1*2*22.4) / (22.4*1*1) = 8 L NH3\n\nHope this helps!
374006
Cu(s) + 2HNO3(aq) = Cu(NO3)2(aq) + H2(g)\n\nCu + 2H+ + 2NO3- = Cu+2 + 2NO3- + H2 (leaving off the s,aq,g)\n\nCu + 2H+ = Cu+2 + H2 (net ionic equation, the NO3-'s just cancel out, they don't change in the reaction).
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NaHCO3(s) + CH3COOH(aq) → CH3COONa(aq) + CO2(g) + H2O(l) \nThis is the real chemical reaction.
359683
4NH3+5O2--->4NO+6H2O\n\n96g O2*(1mole O2 / 32g O2)=3moles O2\n3moles O2*(4moles NH3 / 5moles O2)=2.4 moles NH3\n2.4moles NH3*(1moleNH3 / 17g NH3)=40.8g NH3
428081
It is simply sum of products minus sum of reactants\n\n[H2O + N2] - [O2 + NH3]\n\nSince this is a 1 to 1 ratio reaction, the coefficients are one. (if there was a coefficient for NH3, you must multiply the enthalpy value of NH3 by that # because that is how many moles you have, and the value given is for only one mole.)
464918
C3H6 +  9/2 O2 ---> 3 CO2 + 3 H2O\nassumed oxygen is ideal gas, so\nPV=nRT\nR = 82.057 cm3.atm/gmol.K\nV = 15 l = 15000 cm3 ; T = 37 C = 310 K\nn (mol) O2 = PV/RT = 2.4*15000/82.057*310 = 1.415 gmol\nfrom balanced reaction CO2 = 3/(9/2) * 1.415 gmol = 0.943 gmol\n                                         = 0.943 * 44 = 41.507 gram
473444
The first reaction is simply wrong.  Nitrous acid does not ionize to nitrate.\n\nThe second reaction is a precipitation reaction, showing the net ionic equation.  The compound is ionic.\n\nThe third reaction is an example of SN2 displacement, not elimination.  Only single bonds are formed.
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here, you need to apply the concept of limiting reagent.\n\nhere, N2 is limiting reagent. \n\nSo, if 28 gm (molar mass of N2)nitrogen would give you 34 gm of (molar masss of 2NH3) ammonia.\n\nSo, 3 gm (given) nitrogen would give you how many?\n\n = (7 * 34)/ 28\n\n = 8.5 gm of ammonia.
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2N2H4 + N2O4 -> 3N2 + 4H2O\n\nn(N2H4) = 3.125 mol\nn(N2O4) = 3.333 mol\n\nAs 2mol N2H4 rx c 1mol N2O4, n(N2H4rxn) = 1.563mol, and hence it is the limiting reactant. n(N2O4) = n(N2) = 1:3\nTherefore, 1.563 * 3 = 4.688 mol of Nitrogen gas, which equates to 131.25g.
463629
I assume the reaction in question is:\n\nA(g) <-> 2B(g)\n\nThe equilibrium constant, Keq, for this reaction is given by:\n\nKeq = ([B]^2)/[A]\n\nwhere the quantities in square brackets mean "the concentration of" the relevant species at equilibrium (or, in the case of gases, one commonly uses equilibrium partial pressures instead of concentrations).\n\nPlugging in the values we are given for the equilibrium concentrations, we have that:\n\nKeq = (0.154^2)/0.381\n\nKeq = 0.0622
168511
Well, let's see:  you're releasing 1000 kg/hr of liquid N2, which equilibrates to 1 atmosphere and 20C (STP).\n\nAt STP, one mole of a gas occupies 22.4 liters;\nyou're releasing 1000 kg/hr = (1000 kg*1000g/kg)/28g/mole = 35,714 moles of N2 per hour, and thus:\n\n22.4 liters/mole * 35,714 moles = 800,000 liters (at STP) per hour
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Ba(NO3)2 + Na2CO3-->BaCO3 +2NaNO3\n\nto do a double displacement follow this formula\n\nAB + CD --> AD + CB
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The reaction NaOCl + NH3 = NH2Cl + NaOH is very easy even cold. The reaction NH2Cl + NH3 + NaOH = N2H4 + NaCl + H2O is not so easy as it requires heating to boiling or close to it.\n\nHere's the link> http://www.sciencemadness.org/talk/post.php?action=reply&fid=2&tid=1128&repquote=20454
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This is just a simple stoichiometry problem.  You have 200g of NH3, your first job is to calculate how many moles of NH3 this is.  From the periodic table, find the atomic weights of N and H to calculate the molar mass of NH3:  \n\nN=14 and H=1, so NH3 = 17.  \n\n200g NH3 X (1 mol NH3 / 17g NH3) = 11.76 mol NH3\n\nOkay, now that we know we're working with 11.76 moles of NH3, we can use the mole ratio from the given chemical equation to figure out how many moles of oxygen are needed.  Look at the coefficients in the equation -- we need 5 moles of O2 for every 4 moles of NH3:\n\n11.76 mol NH3 X (5 mol O2)/(4 mol NH3) = 14.7 mol O2\n\nNow we know we need 14.7 mol O2.  All we have to do now is convert that to grams, again using the atomic mass from the periodic table.  The atomic mass of O is 16, so O2=32  \n\n14.7 mol O2 X 32g/mol = 470.4 g O2\n\nSUMMARY:\nConvert grams ammonia to moles ammonia using atomic weights.\nUse the mole ratio to determine needed moles of O2.\nConvert moles O2 to grams O2 using atomic weights.
434591
The temperature difference is 500 K.  If we add 26/deg for NH3, and subtract 0.5 x 27/deg for N2 and 1.5 x 29/deg for H2, we will net out at -0.5 -4.5 = -5.0 J/K per mole of product, so (if i've got the signs right) we will be at -46,000 -2,500 joules/mole, or -48,500 joules per mole.  It is more straightforward to compute the enthalpy changes separately and add them up, but by saving the multiplication till later I can do the numbers in my head.
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Silver is being oxidized, nitrogen from the nitrate is being reduced.\n\nOxidation half reaction:\nAg --> Ag+  +  e\n\nReduction half reaction:\nNO3- +  4H+ + 3e  -->  NO + 2H2O\n\nNet balanced and reduced equation:\n3Ag +  NO3- +  4H+ --> 3Ag+ + NO + 2H2O