how do u calculate the weight percentage of Al in Al2O3?
537504
54/102 =.53 *100=53%Al
802803
The answer is:\n2Al(HCO3)3  -------> Al2(CO3)3  +  3CO2  +  3H20\n\nThe equation is now fully balanced as you have\n2 moles of Al on each side\n6 moles of H on each side\n6 moles of C on each side\n18 moles of O on each side\n\nHope that helps!
22203
AL2 O3 (Aluminum oxide)
625685
Aluminium has a valanly of 3 Oxgygen has 2\nSo\n4Al + 3O2   -> 2Al2O3
223710
Start by balancing Al, then Cl. Al is already  balanced, so when you balance Cl everything is easy from their. \nYou final amswer will be:\n3HCl + Al(OH)3 ----> 3H2O + AlCl3
937581
OK, first we have to balance the equation.  We've got a compound with 3 O's on the right and one with 2 O's on the left, so first let's get them equal:  2Al2O3 + C --> Al + 3CO2.  Now, the C and Al can be used to balance the rest out:  2Al2O3 + 3C --> 2Al + 3CO2.  That answers your second question (3), and we'll need it for the first too.\n\nNow, we need to figure out how many moles of aluminum are in a gram.  The moloar mass of aluminum is 26.98, so there are 26.98 grams in a mole.  There are therefore 1/26.98 moles in a gram.\n\nNow, we can see from the balanced equation that it takes 3 moles of carbon to make 2 moles of aluminum.  So you will need (3/2)/26.98 moles of carbon.  The molar mass of carbon is 12.01, so you need (3/2)*12.01/26.98 grams of carbon.  Do the math and it works out to 0.67 grams of carbon.  It makes sense that it should be less than a gram, since carbon's molar mass is less than half of aluminum's.\n\nSo your answers are:  0.67g, 3.
350595
C whether this helps You,\nhttp://web.umr.edu/~kpeaslee/classes/Met221/KeyHW2.pdf#search='measuring%20SDAS%20in%20Aluminum%20alloy%20casting'\n\nhttp://www.aluplanet.com/eng/info_eco_doc_PO.asp?Mnu=141&Itm=163&Grp=0&Tgl=162&Doc=3804
613586
For calculating weight you will be needing the specific gravity or bulk density of the material it is made of.\nElse you need a weigh balance.
175516
I not sure what style you learned to use to balance chemical equation but here is one way to do it. \nWe have four elements to balance (Al,H,Cl,O)\nLeave the compounds with aluminum with one of each. (Al done)\nThe simplest compound Al is AlCl3. To match the Cl on both sides put a 3 in front on HCl (Cl done)\nIf you add up the H (hydrogen) on the left side of the equation you'll see 6 of them. Add a 3 in front of the H20 to balance that. (H done)\nDo a quick check and you'll see there are 3 oxygen on each side (0 oxygen)\nAll the elements have been balanced, yielding a final equation of:\n3HCl + Al(OH)3 ---> 3H20 + AlCl3
350981
This isn't too bad.\n\nFirst, find the enthalpy of formation for each of the reactants and products.\n\nFe2O3(cr) = -272 kJ/mol\nCO(g) = -110.525 kJ/mol\nCO2(g) = -393.509 kJ/mol\nFe = 0 kJ/mol\n\nThen add the enthalpies of formation for the reactants and subtract the enthalpies of formation for the products:\n\nFe2O3 + 3CO - (3CO2 + 2Fe)\n\n-272 kJ/mol + 3(-110.525 kJ/mol) - 3(-393.509 kJ/mol) - 2(0) = -603.575 kJ/mol - -1180.53 = -576.952 kJ/mol \n\nThat should do it.
404126
Wt percent:\n\nC atomic weight: 12\nCO3 molecular Wt : 44\n% wt = 12X100/44 = 27.3%
999871
if you know the weight of the metal used and then how much of it was used you could calculate it and the same for all the non metal parts
389126
You don't have to 'get' equivalent weights of substances, its very simple to calculate them.\n\nFirst of all, you need to calculate the molecular wt or know the atomic wt of the compound or element you're dealing with.\n\nThe general formula for the equivalent wt of a compound is \n\nEquivalent weight = Atomic or molecular wt/ x\n\nx has different definitions in different contexts:\n\nIf for an element, x is the valency\n\nEg- E.W. of magnesium = 24/2 = 12\n\nIf for an ion, x is the number of units of charge\n\nEg- E.W. of Al+3 ion = 27/3 = 9\n\nIf for a compund, x is the total amount of formal charge present on the cation/anion, i.e. the number of ions multiplied by the charge on each radical.\n\nEg- For NaCl, the value of x would be 1\n    For MgSO4, the value of x would be 2\n    For AlCl3, the value of x would be 3\n    For Ba3(PO4)2, it would be 6,...\n\nIf for an acid, x is the basicity(no. of ionizable H+ ions)\n\nEg - 1 for HCl, 2 for H2SO4,..\n\nThe same applies for a base, x is the no. of ionizable OH- ions.\n\nNow for KMnO4 (acidified or not),\n\nThe molecular wt is 39+55+64 = 158\n\nThe value of x is 1\nSo the E.W. is 158/1 = 158\n\nHowever, if you are interested in the E.W. of Mn in KMnO4, which is more relevant in redox reactions(using acidified KMnO4), calculate the oxidation state of Mn first. It comes out to be +7.\n\nSo the E.W. is 55/7
910345
The formula would be Al2(CO3)3; the numbers are subscripts. \n\nLearn your charges for the polyatomic ions and for the other elements, it will really help.
843494
Forget the potassium ions - they don't do anything. The aluminium metal dissolves in the potassium hydroxide solution, forming aluminate ions and hydrogen gas.\n\nThe ionic equation is:\n\n2Al + 2OH- + 6H2O ---> 2[Al(OH)4]- + 3H2\n\nTo explain how to get this equation, you need the ideas of half equations, and oxidation and reduction.\n\nFor example, Al + 4OH- ---> [Al(OH)4]- + 3e-\n\nand 4H2O + 4e- ----> 2H2 + 4OH-\n\nNow combine them, getting the number of electrons equal, and cancelling down some hydroxide ions.\n\nIf it's any consolation, this exercise would be beyond most A-level pupils in the UK.
891286
Percentage composition means calculating the percentage (by mass) of an atom in its molecule. For instance, we would say:\n\n"Calcium Hydroxide (Ca(OH)2) is x% Calcium, y% Oxygen, and z% Hydrogen"\n\nTo find the percentage composition, first find the total mass of 1 mole of the molecule:\n\nFor Ca(OH)2, there is 1 Ca atom, 2 O atoms, and 2 H atoms. The atomic masses are as follows:\nCa = 40.08 g/mol. \nO = 16.00 g/mol\nH = 1.01 g/mol\n\nThe atomic mass of 1 mol of Ca(OH)2:\nCa + 2*O + 2*H = 40.08 + 2*16.00 + 2*1.01 = 74.1 g/mol\n\nTo find the percentage of Ca, divide the mass of Ca in the molecule by the total mass of the molecule:\n\n%Ca = 40.08 / 74.1 = 0.54 = 54%\n\nDo the same thing with O and H, but remember that there are TWO molecules of O and H in each molecule of Ca(OH)2:\n\n%O = 32 / 74.1 = 0.43 = 43%\n\n%H = 2.02 / 74.1 = 3%\n\n\nNow, repeat this process with the other four:\n\nNaHCO3 = Na + H + C + 3*O = 22.97 + 1.01 + 12.01 + 3*16 = 83.99 g/mol\n%Na = 22.97 / 83.99 =  0.27 = 27%\n%H = 1.01 / 83.99 = 0.01 = 1%\n%C = 12.01 / 83.99 = 0.14 = 14%\n%O = 48 / 83.99 = 0.57 = 57%\n\nCaO = Ca + O = 40.08 + 16.00 = 56.08 g/mol\n%Ca = 40.08 / 56.08 = 0.71 = 71%\n%O = 16.00 / 56.08 = 0.29 = 29%\n\nCaCO3 = Ca + C + 3*O = 40.08 + 12.01 + 3*16.00 = 100.09 g/mol\n%Ca = 40.08 / 100.09 = 0.40 = 40%\n%C = 12.01 / 100.09 = 0.12 = 12%\n%O = 48.00 / 100.09 = 0.48 = 48%\n\nAl2(SO4)3 = 2*Al + 3*S + 12*O = 2*26.98 + 3*32.07 + 12*16.00 = 342.17 g/mol\n%Al = 53.96 / 342.17 = 0.16 = 16%\n%S = 96.21 / 342.17 = 0.28 = 28%\n%O = 192.00 / 342.17 = 0.56 = 56%\n\nI hope this helps!
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