tips of balancing chemical equations?
960383
The tip I give to my student is to always FINISH with the element that has only one type of atom, like O2 or H2.  This way, when you balance it, you won't disturb any work done previously.
908510
It should be Fe2+ + 2OH- --> Fe(OH)2 since you not only need to balance the elements but you also need to balance the charges.
887756
Basic balancing of chemical reactions. You can always add CO2 O2 and H2O to either side. Start with the Carbon. balance the Oxygen last since it is contained in CO2 and H2O.\n\n1) 2 C6H6 + 15 O2 ==> 12 CO2 + 6 H2O\n2) C12H22011 + 12 O2 ==> 12 CO2 + 11 H20\n3) C3H6O + 4 O2 ==> 3 CO2 + 3 H2O\n4) C4H10O + 6 O2 ==> 4 CO2 + 5 H2O\n5) C4H10O + 6 O2 ==> 4CO2 + 5 H2O
677826
1) is already balanced... to be more precise\n2C2H6 + 7O2 --> 4 C02 + 6H2O\n\n2) Fe2O3 + (3)H2SO4 --> Fe2(SO4)3 + (3)H2O
749125
The unbalanced equation is:\nCH3-OH + O2 (+ heat) ---> H2O + CO2\n\nThe balanced equation would be:\n2CH3-OH + 3O2 (+ heat)  ---> 4H2O + 2CO2
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.
856873
The balanced chemical equation and the molecular weights of the reactants and products.\n\nIn simplest form, assume\n\nXY + AB = XB + AY.\n\nAssume the molecular weight of XY is 50, and of XB 35.\n\nFor every 50 grams (or pounds, tons, or any other measure of mass) of XY you use, you will get 35 grams (or whatever) of XB.\n\nEDIT:\n\nOK. Atomic weights are H=1, S=32, Cl=35.5, Fe=56. Molecular weights are FeS=88, HCl=36.5, H2S=34, FeCl2=127.One gram molecular weight of each is its molecular weight in grams. So if you react 88 grams of FeS with 36.5 x 2 grams of HCl (the balanced equation tells you that you need 2 moles of HCl, so you multiply by 2), you will get 34 grams of H2S and 127 grams of FeCl2. \n\nNotice that the weights balance. You put in 88 + (36.5 X 2) grams, or 161 grams. You got out 127 + 34, or 161 grams.\n\nYou need the mass of any one of those compounds - either reactant or product - to determine all the others. Calculate the moles of the one for which you have the mass, determine the ratio of the moles of that one to the others, multiply by that ratio and then multiply result by molecular weight.
342344
The answer is C.\n\nIn order for the equation to be balanced, you have to make sure there are the same number of each element on both sides of the equation.\n\nIn answer C, the number of each element is equal on both sides. \n\nIn answer A, the number of Sodium (Na) and Hydrogen (H) atoms are not equal on both sides of the equation.\n\nIn answer B, the number of Nitrogen (N) and Hydrogen atoms are not equal on both sides of the equation.\n\nIn answer D, the number of Sodium atoms is equal, but there is one more Sulfur (S) atom on the product (right) side of the equation.
4764
You can't balance it because you don't know what it yields.\n\n\nP.S. This needs to be in the chemistry category. You really must not know much about chemistry and physics.
290716
Unless a equation is properly balanced the amount of product and reactant will not be the same, I suppose this is like a recipe where u get the proper number of ingredients(moles) and mix to together to give the you the proper number of products.
727014
start with the O... they are always -2\nsince there are 2, and the compound must balance...\nPb must be +4\n\nso you get +4 -2
844758
Using the Michaelis-Menten equation.\n\nMore details on teh attached link\n\nhttp://www.le.ac.uk/by/teach/biochemweb/tutorials/michment2print.html\n\nGood luck
824141
Start off with the original equation\n\nCaCO3(s) + HCl(aq) --> CaCl2(aq) + CO2(g) + H2O\n\n\nStart off by making the Ca equal on both sides, which is fortunately already done\n\nThen make the Cl equal on both sides by putting a 2 in front of HCL\nCaCO3(s) + 2HCl --> CaCl2(aq) + CO2(g) + H2O\n\nThen make the C's on both sides equal, which again, is already done\n\nNext make the O's equal on both sides, which yet again is already done.\n\nFinally you make the H's equal on both sides, which is done also. \n\nSo the answer to this whole question is to put a 2 in front of the HCl. (I'm guessing you meant to put CaCl2 instead of CaC12 in the products)\n\nThat was easy.
660043
1.) do your own homework\n2.) this is the computer forum, not high-school chemistry.  Next time post to the right furom.
158793
Okay to answer the stoichiometry problem first...You need to balance the equation first...If you don't know how to balance the equation email me and I will explain.  But the balanced equation is:\n\n2Na3PO4 + 3Ba(NO3)2 ----> Ba3(PO4)2 + 6NaNO3\n\nHere is a basic "map" you need to follow:\n\nGrams of Substance A divided by molar mass of A ===> Moles of A times the coefficients of A and B from balanced equation to get moles of substance B===> Moles of substance B times molar mass of substance B ===> Grams of substance B\n\n\nTo start we need to calculate the molar mass of the two compounds we are mixing together\nNa3PO4 = 151.97g/mol\nBa(NO3)2 - 213.34g/mol\nThen we need to find the molar mass of the product, which is Ba3(PO4)2 - 577.97g/mol\nIf you don't know how to find the molar mass, email me. Otherwise, read a chemistry book.  Anyway the steps are to convert grams to moles then multiply by the coefficients of our balanced equation and then multiply again by grams of \n Ba3(PO4)2...Seems complicated, but it's really not.  \n\n1. To see if Na3PO4 is the limiting reactant we divide 3.50g by it's molar mass of 151.97g/mol to get moles of Na3PO4\n\n3.50/151.97 = 2.30*10^-2 mol. Take this and multiply by the coefficients of our reactant and product:\n\nTake the coefficient in front of Na3PO4 and place that in the denominator  /2 \nTake the coefficient in front of our product Ba3(PO4)2 and put it in the numerator  1/\n\nNow take the calculated moles of \n2.30*10^-2 and multiply them by (1/2) <----  the coefficients of reactant and product\n\nThis gives you moles of Ba3(PO4)2\n(2.30*10^-2) * (1/2) = 1.15*10^-2 mol\nNow multiply this number by the mol weight of our product which is 577.97g/mol and that gives you grams of product produced.\n\n(1.15*10^-2) * 577.97 = 6.66g of Ba3(PO4)2\n\nWhew!\n\nNow on to the next one....I won't explain fully as it's exactly the same procedure.\n\n6.40g/213.34g/mol = 2.99*10^-2 mol of Ba(NO3)2 (reactant)\n\n(2.99*10^-2) * (1/3) = 9.99*10^-3 mol of Ba(PO4)2 (product) \nI used (1/3) because the number in front of the reactant is 3 (goes in the denominator) and the 1 is the coefficient in front of the product.\n\nNow   (9.99*10^-3) * (577.97) = 5.78g of Ba3(PO4)2\n\nokay so upon inspection, the limiting reactant is Ba(NO3)2 and it yields 5.78 grams...\n\nYou ask if 3.50 grams are actually yielded, what is the %\n\nSo....(actual yield/theoretical yield)*100 = % \n\n(3.50/5.78)*100 = 60.55%\n\nAlrighty....Now onto problem #2\n\nBalance: Al(NO3)3 + Na2S ----> Al2S3 + NaNO3      \n\n(A) Al(NO3)3 + (B) Na2S ----> (C) Al2S3 + (D) NaNO3\n\nPut variables in front of each reactant and product like I did above.  Now assign an element to equal the variable on both sides.  In our case:\n\nAl  -    A = 2C\nNO3 -   3A = D\nNa  -   2B = D\nS   -   B = 3C\n\nNow make one letter equal one. We'll make A = 1\nSo..\nA = 1\nB = 3/2  (by algebra)\nC = 1/2  (by algebra)  (email me if you don't know)\nD = 3\n\nwe can't have fractions in there, so we multiply all of them by two to get the 2 in the denominators out, and we end up with\n\nA = 2\nB = 3\nC = 1\nD = 6    With a final equation balanced out to look like:\n\n2Al(NO3)3 + 3Na2S ---> Al2S3 + 6NaNO3\n\nHope all this was clear!!!  Enjoy
1040448
ask your chem prof or tutor ..it's hard to explain over the internet!
881744
First, balance the equation.\n\n2LiOH + CO2------>Li2CO3  +  H2O\n\nSo, for every 2 LiOH, there will be 1 Li2CO3 and 1 H2O\n\nSo, for every 5.4 LiOH, there will be 2.7 Li2CO3 and 2.7 H20