In the decomposition equation below how many grams of Oxygen is made when 15grms of H2O2decomposes.?
915044
7.04 grams\n\n15 grams of water is .44 mol and according to the equation, the ratio between water and oxygen is2:1 so if .44 mol of water is decomposed. then .22 mol of oxygen will be produced. and .22 mol of oxygen is .22 *32 =7.04grams
833745
first you have to balance the equation:\n2SO2+O2=2SO3\n\n1) 96gO2*(1moleO2/32g O2)= 3mole O2\n2) 3mole O2*(2moles SO3/1mole O2)=6mole SO3\n3) 6moles SO3*(80.1g SO3/1mole SO3)=480g SO3
863022
Here you are:\n\nFe+2 + 1/2O2 (g) + H2O (l)  ---> Fe (OH)2 (s)\n\nFe (OH)2 (s) + 1/2O2 (g) ---> Fe2O3 (s) + 2H2O (l)
952215
[H3O+][OH-] = 1.0 x 10^-14 M always at 25C\nso if [H3O+]= 5.0 x 10^-10M\n[OH-]= 2.0x10^-5
470703
reduced: 2H2O + 2e- ---> H2 + 2OH-\n\noxidized: 2I- ---> I2 +2e-
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)
866210
2NO+O2--->2NO2\n\nCaO+H2O--->Ca(OH)2\n\nCaCO3--->CaO+CO2
215604
C6H12O6(s) ---H2SO4--> 6C(s) + 6H2O(in H2SO4 solu.) + Heat
789547
first you need to balance the equation\n\nC3H8(g) + O2(g) -------> CO2(g) + H2O(l) (unbalanced)\n\nCompare carbon atoms on both and you will make three CO2 for every one C3H8.  That leaves 8 Hydrogen atoms to make water which has 2 hydrogen atoms per molecule so you would need  4 H2O to balance that out. The equations starts to look balanced like this\n\nC3H8(g) + O2(g) -------> 3CO2(g) + 4H2O(l) (unbalanced)\n\nnow we have to figure out how much oxygen is needed to balance the whole thing.  There are 2 Oxygens in CO2 and 1 in H2O so we have a total of (3x2)+(4x1)=10 oxygen on the right side of the equation. since O2 has 2 oxygen we need 5 of them to get 10 so the equation looks like this.\n\nC3H8(g) + 5O2(g) -------> 3CO2(g) + 4H2O(l) (balanced)\n\nsince oxygen has an atomic mass of 16.000 amu then O2 has a mass of twice that or 32.000 amu=32g/mol\n\nfrom our equation we get that there is 1 mole of C3H8 to every 5 moles of O2. \n\n5molx32g/mol=160g
702491
Mn2+ +2NH3 +2H2O  <-->  Mn(OH)2 + 2NH4+ (+:sup, 4:sub)\n4Mn(OH)2 + O2 + 2H2O  <-->  4Mn(OH)3\n2Mn(OH)2 + O2  <-->  2(MnO2 . H2O)\n\nMn(OH)2 white sediment\nMn(OH)3 brown sediment\nThe oxidization is faster if we add H2O2
116913
I think it is this:\n\n            \nC12H22O11(s) --- HNO3---->  12C(s) = 11H2O(g) + heat
651011
H2O2 + OC(NH2)2 -> HCOOH + NH4OH -> NH4(HCOO)
222097
OH-(aq) + HC2H3O2(aq)---> H2O(l) + C2H3O2-(aq)
761497
1) The first one is wrong on the last person who answered, O2 weighs 32g/mol because there are 2 O's in it, 16 each.  So in 1.33x10^4 (13,300mol) you have 4.256x10^5g of O2 (425.6kg).\n\n2) Recognize that hydrogen, oxygen, and chlorine are diatomic gasses, so they always occur in pairs (H2, O2, Cl2).  Others like this are nitrogen (N2), fluorine (F2), bromine (liquid @ room temp. Br2) and Iodine (solid at room temp, I2).  So your equations would be:\n\n2H2+O2-->2H2O\nH2+Cl2-->2HCl\n\n3) PV=nRT and solve for n, number of moles.  Put everything in correct units, liters, atmospheres, moles, and kelvins, and R is a constant R=0.0821 L*atm/mol K.  Kelvin temp=Celcius temp +273\n\nSolving for n:\nn=PV/RT=(1.09atm)(0.942L)/(0.0821L*atm/mol*K)(287K)\nn=0.0436 mol\n\nSo, when we usually want to know a number of moles given a mass and having a molar mass we can look up, we say,\nn=mass/molar mass\n\nin this case, we need to find molar mass, so:\nmass/n=molar mass\n3.17g/0.0436mol=72.71g/mol
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.
823190
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!!!
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......