i hate balancing equations can some one help me plz?: P4 O10 + H2O = H3 PO4?
957414
P4 O 10 + 6 * H2O = 4 H3 PO4
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
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......
812798
3Ca(OH)2+2H3PO4>Ca3(PO4)2+6H2O
832115
N2H4 + H2O -> N2H5+ + OH-\n\nKb = c[N2H5+] * c[OH-] / c[N2H4]\n\nAssuming that c[N2H5+] = c[OH-] and that c[N2H4] >> c[OH-] (so that we can use the starting concentration for c[N2H4]), we get approximately\n\nc[OH-] = sqrt( Kb * c[N2H4])\n\nor logarithmized\n\npOH = 1/2 ( pKb + 10^-(c[N2H4])\n\npH= 14 - pOH\n\n\ncalculator time.
225876
Balancing Chemical Equations is absolutely essential if you want to determine quantities of reactants or products. An unbalanced Chemical Equation gives only the identify of the beginning reactants and the final products using the appropriate formulas as well as the conditions of temperature, physical state, and pressure conditions under which the reaction is to operate under. However an unbalanced equation can say nothing about the quantities involved until the equation has been balanced. A balanced equation assures that the Conservation Law of matter is obeyed. The total mass of reactants must equal the total mass of products. A balanced equation is like a recipe. It tells you the proportional quantities of each substance involved. The coefficients that appear to the left of each formula can be interpreted as mole units or molecules, but for practical purposes, the coefficients will be interpreted as mole units. No coefficient is interpreted as 1 mole. The following principles should be employed when balancing a Chemical Equation by inspection:\n\n   1. Never touch subscripts when balancing equations since that will change the composition and therefore the substance itself.\n   2. Check to be sure that you have included all sources of a particular element that you are balancing on a particular side since there may be two or more compounds that contain the same element on a given side of an equation.\n   3. I would suggest that you adjust the coefficient of mono atomic elements near the end of the balancing act since any change in their coefficient will not affect the balance of other elements\n   4. When there are a group of atoms that are acting as a unit such as a polyatomic ion and they appear intact on both sides of the equation, it is best to balance them as a self contaned group. For example, if there are Phosphate groups, PO4-3, that appear on both sides, balance the phosphates as a group instead of separating the Phosphorus and Oxygens. It can be done either way, but there is less likely of a mistake if they are balanced as Phosphate groups.\n\nConsider the example of:\n\n(NH4)2CO3 ---> NH3 + CO2 + H2O\n\nThe first thing to do is choose a starting point. If we choose the carbon as the starting point, we go nowhere, since there is already one carbon on each side. Choosing oxygen is not good, because the oxygen is divided between two species on the right side. The interesting part is in the ammonium (NH4) ion. If we start with the nitrogen on the left (in the ammonium ion), we see that there are two nitrogens and therefore we must put at "2" in front of the ammonia on the right side.\n\n(NH4)2CO3 ---> 2 NH3 + CO2 + H2O\n\nSince we are already working with the ammonia, now consider the hydrogens that make up the rest of the molecule: the six hydrogens in the ammonia plus the two hydrogens in the water make eight hydrogens. Bouncing back across to the left side, we count eight hydrogens; since no other atomic species have been changed, we can count up everything and see that the reaction is now balanced:\n\n2 N, 8 H, 1 C, and 3 O on each side.\n\nNotice how the reaction itself led us through the steps of the balancing process. A more complex reaction would have involved more steps, but the process would have been the same.\n\nThis "inspection method" is the first one tried by chemists in the real world. Only when it fails to work, as in the case of many oxidation-reduction reactions, for example, are more complex procedures used.\n\nHere is another example:\n\nFe2(SO4)3 + K(SCN) ----> K3Fe(SCN)6 + K2SO4\n\nK3Fe(SCN)6 is pretty complicated. Now look at it and see what ions are in it that are not shared on the same side of the reaction. These are SCN and Fe.\n\nWe start to balance these\n\nFe2(SO4)3 + 6 K(SCN) ---> K3Fe(SCN)6 + K2SO4 Now we go left to right.\n\nThe irons are messed up, two on the Left, and one on the right. We adjust\n\nFe2(SO4)3 + 6 K(SCN) -- 2 K3Fe(SCN)6 + K2SO4 .\n\nNow t
970544
Here is what you do: \nDo the same operations as I did on the equations with the H values and you will be OK (Do not forget to use the same multipliers on the H as I did on the reactions).\n\n10PCl3 + 5O2 ---> 10Cl3PO\n10PCl2 + 10Cl2 --- 10PCl5\nSubstract the equations to get (keep coeff positives by moving to the other side):\n\n5O2 + 10PCl5 ---> 10Cl3PO + 10Cl2\nP4 + 5O2 ---> P4O10\nSubstract and get\n10PCl5 + P4O10 ---> P4 + 10Cl3PO + 10Cl2\nP4 + 6Cl2 ---> 4PCl3\n\nSum the two to get (P4 goes):\n\n10PCl5 + P4O10 ---> 10Cl3PO + 4Cl2 + 4PCl3\n\n4 PCl3 + 4Cl2 ---> 4PCl5\n\nSum the two equations\n\n10PCl5 + 6P4O10 ---> 10 Cl3PO.
861843
Start with the Ca then balance the number of PO4 ions, then balance the Oxygen and the hydrogens should balance themselves out.  Its just trial and error. Monkey with it until it works.
378344
At equilibrium, we have this below relationship between K and the concentration of all species.\n     \n　　　　　　　　　　　　　[H2]^2 [O2]\n　　　　　　　　　　　K= ----------------\n　　　　　　　　　　　　　　[H2O]^2\n\nSo we could get the answer by solving this equation.
979796
2H3PO4+6NaOH-------->6H2O+2Na3PO4\n\nThere has to be two moles of phosphoric acid and six moles of sodium hydroxide for it to balance out
780875
a)4 Fe + 3O2--->2 Fe2O3\n\nCH4 + 2O2 ---> CO2 +2 H2O
957905
the balanced reaction\nC12H22O11 + 12 O2 ---> 12 CO2 + 11 H2O\nso the right answer are:\nc. \nand\ne.
151775
giving some sample equations would help!\n\nyou have to make the left hand side balance wilh the right hand side...\n\nsuppose you have the unbalanced equation\n\nC4H10 + O2   =    CO2  +  H20\n\nin this equations the easiest thing to balance first would be the carbon so on the right hand side you put \n\nC4H10  + O2   =   4CO2  +  H2O\n\nThen the hydrogen,you have 10 Hydrogens on the left so you need 10 on the right\n\nC4H10  +  O2  =  4CO2  + 5H2O\n\nthus you will need 13 oxegen atoms on the left side.. since oxegen exists as O2 and 13 is an odd number you multiply everything by a factor of 2 to make it just right... so your final equation looks like....\n\n2C4H10  +  13O2  =  8CO2   +  10H2O
105211
Pressure * volume = number of mol * temperature (in ^oK) * 8.314\n\nYou have all the information you need right there. \nFirst solve for n for the initial state. \nSince you haven't added any more material you can then use the same value of n for the second state and solve for P.\n\nThis is sixth grade algebra. Anyone being asked to solve gas equations had better be able to do it standing on their head or they are in real trouble.
157730
I think FePO4 +O2 >>>>> Fe2O3 + PO2. balance it and solve it . n its simple stoichiometry.
980488
Hmm. Maybe all you need is the guideline to balance it?\n\nFor beginners, this is probably the best way. After some time, you can omit the unnecessary steps.\n\nFirst, list down the number of atoms of elements present on both sides of equation: reactants vs products. In this case, 1S,4O vs 1S,3O.\n\nSecond, find a common number of atom(s) of an element on either side. Knowing that there is 1S on both sides, the reacting O2 molecule will thus have a stoichiometric coefficient of 1/2. This will ensure 3O on either side.\n\nThird, multiply throughout with 2 since since fractions are not preferred.\n\nOf course, math enthusiasts usually get the hang of this faster by looking for common factor and this method is exceptionally effective (faster) when reactions become complex.\n\nBut the best way is still, of course, to practice XD
410035
good news...10.63 is correct. Now how to get there.  \nMorp + H20 ---> Morp+ + OH-\ninitial concentration of Morp=.115M  everything else is 0.\nEquilibrium concentration of Morp=.115M-X, H20 ignore, Morp+=X, OH-=X.   Set up your equilibrium equation.  Kb=[OH-][Morp+]/[Morp]  pKb=-log Kb  so Kb=1.62e-6.  1.62e-6=[X][X]/[.115-X]  We can ignore the X subtracted from .115 since it is very small when compared.  so solving for x we finally get X=4.32e-4 which is the OH- concentration.  pOH=-log OH-  pOH = 3.36.  pH = 14-pOH .....pH=10.63.