How many grams of nitric acid, HNO3, can be prepared from the reaction of 92.0 g of NO2 with 36.0 g H2O?
924787
First of all\n3 moles of NO2 give you 2 moles of HNO3, while 1 mol of water gives you the same amount of product. We must find out which reactant gives you the least product, and that will be your maximum amount of product.\n36,0g H2O*(1mol H2O/18g H2O)*(2moles HNO3/1mol H2O)*(63g HNO3/1mol HNO3)= 252 g of HNO3\n\nNow with NO2\n92,0g NO2*(1mol NO2/46g NO2)*(2 moles HNO3/ 3moles NO2)*(63g HNO3/1mol HNO3)= 84,0g HNO3\nTherefore, the biggest amount of HNO3 you can expect to get is 84,0g.
262778
73.93 grams\nFirst you use the Ideal Gas Law: PV=nRT >P is pressure in atm., V is volume, n is moles, R is the constant: .0821, T is the temperature in Kelvins> (1atm.)(3.16L)=(n)(.0821)(305K) > .126195563 moles. Now you use a proportion. 9.33grams/.126195563moles = x grams/ 1 mole > 73.93 grams
959167
HNO2...all others are strong acids.
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%
876773
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.
294143
add:\n\nH2S2O8(ag)+ NH3(aq) + ? -->(NH4)2S208(aq)  + ?\n\nAmmonia can be prepared in the laboratory by treating ammonium chloride with sodium hydroxide:\n\nthe net ionic equation to this reaction is:\n\n NH4+(aq) +  OH-(aq) ---> NH3(g) + H2O(l) in system 1\n \nand\n\n NH3(g)   = NH3(aq) in solution in system 2\n\nyou finish it ,if you need more help just e-mail me
929071
2C18H18 + 45 O2 --->  36 CO2 + 18 H2O\n3Cu + 8HNO3 ----> 3Cu(NO3)2 + 2NO + 4H2O
904934
2HNO3 + Mg(OH)2 -->Mg(NO3)2 +2H20\n\n2H+(aq) + 2NO3-(aq) + Mg+(aq) + 2(OH)-(aq) -->Mg+(aq) + 2(NO3)-(aq) +2H20(l)\n\n2+(aq) + 2(OH)-(aq) --> 2H20(l)
850425
KOH + HNO3 ---> KNO3+ H20\n\nmoles of KOH = 0.025L*0.2mol/L = 0.005 mol\nconcentration of nitric acid solution = 0.005mol/0.030L = 0.167M\nconcentration of KNO3 = 0.005mol/0.055L = 0.091M
962694
Tehe answer is:\nAg + HNO3 -> AgNO3 + H\nAgNO3 + KBr -> AgBr + KNO3
981241
HNO3 is synthesized from ammonia.\n\n4NH3 + 5O2 -> 4NO + 6H2O\n \nthen:\n\n2NO + O2 -> NO2\n and then NO2 is dissolved in water:\n(the prevailing reaction is:) \n3NO2 + H2O -> 2HNO3 + NO\n\nNO is recycled back to teh previous stage.\n\nThose reactions are oversimplified, while in actuality, there're many other side and intermediary reactions
215729
K2IC4H6 + HNO3 ---  K2HI + NO3C4H6
98695
My answer is,\n \nNH4NO3(ag) + KOH(aq) -> NH4OH(aq) + KNO3(aq)
765235
Here you are:\n1. multiply the first equation by -1 to get\nN2 + O2 ---> 2NO  H= 180 kJ   {1}\n2. Sum the above equation with the second one:\n2NO + O2 ---> 2NO2 H= -112 kJ   {2}\n\nThe sum is\nN2 + 2O2 ---> 2NO2  (This is what you need)\nSo now sum the Hs for the rearenged equation {1} and the second equation {2}.  This will be the H for the equation you need\n\n-112 + 180 = 68 kJ.
700367
1) Balance the reaction equation:\nPb(NO3)2  + 2 LiBr -> PbBr2 + 2LiNO3\n\n2) Calculate the amounts we start with\nPb(NO3)2 :  2.05 L * 0.0550 mol/L = 0.1128 mol\nLiBr : 1.52 L * 0.0618 mol/L = 0.0939 mol\nAs we need 2 moles of the latter for 1 mole of the former, the reaction is limited by the amount of LiBr\n\n3) Calculate the amount of PbBr2 produced\n0.0939 mol of LiBr will react with 0.470 mol of Pb(NO3)2 to produce 0.0470 mol of PbBr2. This is 0.0470 * (207.2 + 2* 79.9) =17.249 gram\n\n4) Calculate the yield\nThe yield is 15.784 / 17.249 = 91.51%
866626
O3 = 16g x 3 = 48g/mol.\n 1 Mole O3 = 48g  (÷ 4)\n 0∙02083333...Mole = 1g    (x 0∙740)\n 0∙014516666.Mole = 0∙740g\n\nNO = 14g + 16g = 30g/Mol.\n 1 Mole NO = 30g\n 0∙03333...Mole = 1g\n 0∙022333.. Mole = 0∙670g\n\n If the O3 splits to give O2 + O,\nthen the reaction is 1:1, and O3 is the limiting reactant.\n\nExcess NO: \n = 0∙022333.. Mole of NO  -  0∙014516666.Mole of O3\n = 0∙0068666...Moles of NO.\n\n If the O3 splits to give O2 + O, and splits again to give O + O + O, then the reaction is 3:1, and NO is the limiting reactant.\n\nExcess O3:\n = 2/3 x  0∙014516666.Mole \n = 0∙009677777....Moles of O3.
840582
1.H2O=18g\n\n2.CH3Cl=52g\n\n3.SO4=96 g\n\n4.NH3=17 g\n\n5.C3H7O=59 g\n\n6.NO2=46g