For the reaction NH3 + O2 → NO + H2O?
825261
First you must balance the equation:\n4NH3 + 5O2  -> 4NO + 6H2O\nthere are 4N, 12H and 10O on each side\nyou have 4.78 moles of O2 you need to divide it by the number of moles of O2 that react and multiply it by the number of moles of NO produced.\n4.78 / 5 x 4 = 3.824moles\nHope this helps
756485
The water isn't involved in the reaction, so take it out.  Then, just put a 2 in front of the Nh2(SO3)H and a 2 in front of HNO3, and you're finished.\n\nNow, go take a nap or go to sleep...
814611
CdSo4+NH4OH ---->Cd(OH)2 + (NH4)2SO4
727004
(NH4)2Cr2O7 -> Cr2O3 + 4*(H2O) + N2\n\n(NH4)2SO4 + 2*(NaOH) -> 2*(NH3) + 2*(H2O) + Na2SO4\n\nFe2(SO4)3 + 6*(NH3) + 6*(H2O) -> 2*(Fe(OH)3) + 3*((NH4)2SO4)
412530
1) H3COCH3 (or C2H6O) + 3O2 (g) --> 2CO2 (g) + 3H2O (g)\n2) HNO2 (aq) + NaOH (aq) <--> NaNO2 (aq) + H2O (l)\n3) 2K+ + Cr2O7(2-) + Fe(NO3)2 + 2H+ <--> CrO4(-) + Fe3+ + 2K+ + 2NO3(-) \n4) Na3PO4 + NH4NO3 = No reaction, all soluble salts.\n5) 2HCl (aq) + Na2S (s) --> 2NaCl (aq) + H2S (g)
712891
In one experiment 32.40 g of NO is mixed with 19.20 g of O2.\n\n\nThe reaction goes: 2NO + O2 -> 2NO2\n\n\nFirst, you have to find out which is the limiting reactant.\n\nno of moles of NO\n= mass / molar mass\n= 32.40 / (14 + 16)\n= 32.40 / 30\n= 1.08\n\nno of moles of O2\n= mass / molar mass\n= 19.20 / (2 x 16)\n= 0.6\n\nGoing by the equation, there must be twice the no of moles of NO than O2. However, the no of moles of O2 supplied is more than half of the no of moles of NO. Thus, O2 is in excess and NO is the limiting reactant.\n\nWorking with the no of moles of NO,\nno of moles of NO2\n= no of moles of NO\n= 1.08\n\nmass of NO2\n= (no of moles) x (molar mass of NO2)\n= 1.08 x (14 + 2 x 16)\n= 1.08 x 46\n= 49.68g\n\nAnd there you have it! :D
565135
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
866210
2NO+O2--->2NO2\n\nCaO+H2O--->Ca(OH)2\n\nCaCO3--->CaO+CO2
766197
CO2\nCarbon Dioxide\nHere is the reaction:\nHC2H3O2(aq) + NaHCO3(aq) ----> NaC2H3O2(aq) + H2O(l) + CO2(g)
1018313
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
840588
First of all, you need a chemical equation,\n\nNO + O2 ----> NO2\n\nThen, balance it,\n\nNO + 1/2 O2 -----> NO2\n\nNow, what this balanced equation tells you is that the mole ratio between NO and O2 is 2:1. i.e., you need 2 parts of NO to react with every 1 part of O2.\n\nNow, you need to calculate the total number of moles of NO and O2 respectively,\n\n# mole of NO = 17.26/(14+16) = 0.575 mole\n\n# mole of O2 = 13.15/(16+16) = 0.411 mole\n\nAs said before, for every 1 mole of O2, it takes 2 moles of NO to react. From the number of moles of NO and O2, it is clear to see that NO is the limiting reagent. Why? Because if you react all 0.575 mole of NO, it takes only 0.575/2 mole (0.2875 mole) of O2, which you have a lot more than that.\n\nThe limiting reagent is the substance that is NOT in excess, which then determines how much product will be produced (because once this substance is all gone, there will be no reactants to react and give product).\n\nNow, from the balanced equation, you know the mole ratio between NO and NO2 is 1:1. Therefore, the number of moles of NO2 that be formed will be 0.573 moles.\n\nMass of NO2 = # mole * molecular weight\n= 0.573 * (14 + 16 + 16) = 26.465 g
962694
Tehe answer is:\nAg + HNO3 -> AgNO3 + H\nAgNO3 + KBr -> AgBr + KNO3
136146
The 1st answer is wrong. The reaction is\n\nNa2CO3(aq) + SrCl2 (aq)--> 2NaCl (aq) +SrCO3 (s)
708721
Here is the right equation for nitric acid and sodium bicarbonate\n\n\nHNO3 + NaHCO3 ---> NaNO3 + H2O + CO2\n\nIf you do not like the sodium:\n\nHNO3 + HCO3(-) --> NO3(-) + H2O + CO2
809514
Here you are:\n\nFirst write the combustion equation and balance it:\n\nH3NNH2(l) + O2(g) ---> N2(g) + 2H2O(g).\nThe heat will be the diffrence of enthalpies of formation between the products and the reactants. Do not forget to multiply by the stoichiometric coefficient.\n\nRemember that the enthalpies of formation of O2 and N2 are 0.\n\nYou will also need to find the enthalpy of formation of H2O(g) in the table of your book.  It is -241.8 kJ/mol\n\nHf(H2NNH2) = [2*(-241.8) + 1*0] - [1*50.6 + 1*0]\n\nHf(H2NNH2) = -483.6 - 50.6 = - 534.2 kJ/mol
605005
NaHCO3(s) + CH3COOH(aq) → CH3COONa(aq) + CO2(g) + H2O(l) \nThis is the real chemical reaction.
764101
In all reactions one uses that 2 OH -> H2O + O, the split of oxygen goes to create the ion. Finally balance the equation:\n \nSO3 + 2 NaOH -> Na2SO4 + H2O\n\nCO2 + 2 KOH -> K2CO3 + H2O \n\nSO2 + Ba(OH)2 -> BaSO3 + H2O