what is the symbol equation of iron (III) oxide?
918374
Fe2O3 -- Fe+O2\n\nthis is the balanced form\n\n2(Fe2O3)--- 4Fe+3O2\nwithout the parentheses
802786
I'll show you how to do one; then you'll know how to do the rest:\n\na. On the periodic table, find the atomic weight for both Fe and O.\nFe = 55.845, O = 15.9994\nMultiply by the number of atoms of each in each molecule of Fe2O3 to get the molecular mass (the mass, in grams, of one mole of the compound) of Fe2O3:\n2(55.845) + 3(15.9994) = 159.6882\nDivide 150 by this number to get the number of moles:\n150/159.6882 = 0.939 moles
507416
I2 + H2O = IO3(-) + H+\n\nI2 = I-
941802
In order to do this, you need to realize what the balanced equation is really saying, and what relation that has to mass.  Your equation says that for every 4 iron atoms that react with 3 diatomic oxygen molecules, you'll produce 2 iron (III) oxide molecules.  Molecules are great and all, but that doesn't help you when you're trying to figure things out in terms of grams.  That's where Avogadro's number and the concept of the mole come into play.\n\nOn the periodic table, you'll notice that there are two numbers for each element:  the atomic number, which is the number of protons in an atom of the element; and the atomic weight (or mass), which is the total weight, in grams, of one mole (6.022 X 10^23 atoms) of that element.  (The atomic weight is also roughly equal to the total number of protons and neutrons in the most common isotope of the element.)\n\nIn order to figure out how much product you'll produce, you need to figure out how many moles of Fe you have, then work out the proportions.  Start by figuring the molar mass of each molecule in the reaction:\nFe = 55.845\nO2 = 2(15.9994) = 31.9988\nFe2O3 = 2(55.945) + 3(15.9994) = 159.8882\n\nNow figure out how many moles of Fe you have:\n4086 g/55.845 g/mol = 73.167 mol\n73.167 mol/4 mol Fe\n\nNow figure out how many moles of Fe2O3 you produce:\n(73.167/4) X 2 = 36.584 mol Fe2O3\n\nMultiply this by the molar mass of Fe2O3 to get the mass of the product:\n36.584(159.8882) = 5849.350 g Fe2O3\n\nb) Do the same thing you did in a), just substitute O2 for Fe2O3:\n(73.167/4) X 3 = 54.875 mol O2\n54.875 x 31.9988 = 1755.942 g O2
808775
Na2C2O4\n Na +1, C +3, O -2
697397
NaOH\nSn(NO3)2\nCo2O3\nFeF2\nCuS2\nFe(OH)3\nAgBr
825401
First of all I assume  Na2Co3 should be Na2CO3.\n\nFe(CO)5 + 4NaOH = Na2Fe(CO)4 + Na2CO3+ 2H2O\n\nFirst, balance out the metals (Na and Fe) then C, which was already balanced. Then I chose to balance out H and which automatically balanced out O.
1125539
343.2 g of Fe2O3 - take the molar mass of Fe2O3. 2 irons is 55.85 * 2 = 111.70 grams per mole. 3 oxygens is 16 * 3 = 48. Divide 343.2 g by 159.70 g/mol and you get 2.15 moles.\n\nYour original reaction uses 2 moles of ferric oxide, so you need to multiply everything by (2.15 / 2) to get your ferric oxide to say 2.15 moles. When you do this, you will change your Kcal value to the right number.\n\n2 Fe2O3 (s) + 196.5 Kcal --> 4 Fe(s) + 3 O2 (g)\n2.15 Fe2O3 (s) + 211.2 Kcal --> 4.30 Fe(s) + 3.23 O2 (g)\n\nNote that I really didn't care about significant figures, so you should probably make sure I didn't round off incorrectly. The 2 and 4 and 3 in the equation are exact (2.0000....), but the 196.5 is an inexact number.\n\nThe second equation is easier, because you already have 0.393 in moles. You need to multiply by something that turns a 2 C3H6O into 0.393 C3H6O. So that's simply (0.393 / 2).\n2 C3H6O + 8 O2(g) --> 6 CO2 (g) + 6 H2O(g) + 853.6 kcal \n0.393 C3H6O + 1.572 O2(g) --> 1.179 CO2 (g) + 1.179 H2O(g) + 167.7324 kcal\n\nSo basically, you need to find some known number of moles, and multiply and divide your balanced equation until it has that number of moles in the equation. Then you read off whatever you need.
590452
(NH4)2 SO4  + (FeSO4)2 + 2H2O____(\nNH4)2SO4Fe2(SO4)3.24H2O\n\nAmmonium ferric sulphate
748280
PH3:                           Phosphine\nAg2SO3:                    Silver Phosphite\nFeCO3:                      Iron(II) Carbonate\nN2O4:                        Dinitrogen tetroxide\nPb(OH)2:                   Lead(II) Hydroxide
806049
A nonstoichiometric compound is anything in which the elements do not occur in proportions of whole numbers.  Usually these are solids which have some defect in their crystal lattice structure, meaning it is missing some ions.\n\nA common example is of the three biggies in the iron oxides.  Fe2O3 and Fe3O4 are typically stoichiometric... if you have 2 million Fe ions in your Fe2O3, you will have very near 3 million O ions.  FeO, on the other hand, is often slightly short of iron, and is written Fe(1-x)O, in which x is some small number.
734376
Since the equation you listed isn't balanced, I'm going to attempt that first:\n\nFe2O3 + 3(CO) ----> 2Fe + 3(CO2)\n\nsince 1 mole of Fe2O3 seems to produce 2 moles of Fe, it would appear we need 5.9/2 = 2.95 moles of Fe2O3\n\nGiven that 1 mole of these elements weighs: (see source)\nO = 16g\nFe = 56g\n\n2.95 moles * ( 2 * 56g + 3 * 16g ) = 2.95 * 160g = 472g\n\nHope that helps.
606075
Cu(s) + 4HNO3 (aq)--> Cu(NO3)2(aq) + 2H2O(l) +2NO2(g)
702529
Name: Iron \nSymbol: Fe \nAtomic Number: 26 \nAtomic Mass: 55.845 amu \nMelting Point: 1535.0 °C (1808.15 K, 2795.0 °F) \nBoiling Point: 2750.0 °C (3023.15 K, 4982.0 °F) \nNumber of Protons/Electrons: 26 \nNumber of Neutrons: 30 \nClassification: Transition Metal \nCrystal Structure: Cubic \nDensity @ 293 K: 7.86 g/cm3 \nColor: Silvery
777814
Q1)\nCa3 (PO4)2  (MW)molecular weight = 3*40 + 2 * (95) = 310 the oxygen weight is 16*4 = 64 .so % of O is (64/310) *100 = 20.6%\n\nQ2)\nSO4 (NH4)2 ---> MW = 32+4*16+ 2*18 =132 Mass of sulfur 32 \nhence  % S = 100*32/132 =24.2\n\nQ3) \n\nFe2 (SO4)3 ----> MW = 56*2+96*3=400 \n\n% of iron 100*112/400 = 28%
537646
The thing is that iron III hydroxide isn't soluble in water.  It just sits there and doesn't make many hydroxide ions, which are necessary for a strongly basic solution.  Barium hydroxide, however, is soluble and breaks up into barium ions and hydroxide ions, giving a strongly basic solution.
864677
carbon monoxide = CO\nIron (II) Chlorate= Fe(ClO3)2\nLead (IV) oxide = PbO4\nNitric acid = HNO3\nSulfuric Acid = H2SO4\n\nall numbers are subscripts.