Please Write a balanced chemical equation for the double replacement reaction?
957775
Ca(OH)2 + CO2 --> C(OH) + CaO\n\nIs unbalanced..
502725
That second reaction should be Pb(s) + Cl2(g) ---> PbCl2(s).  Reverse the second reaction and then add the two together and you will get the resultant reaction.\n\nPbCl2(s) ---> Pb(s) + Cl2(g)    H = +359.4 kJ\nPb(s) + 2Cl2(g) ---> PbCl4(s)   H = -329.2 kJ\n\nPbCl2(s) + Cl2(g) ---> PbCl4(s)  H = +30.2 kJ
119628
So, usually the best way to do these is to just fiddle with them until you get them right, but for the sake of it, let's go with a systematic approach using algebra.  First we assign a variable in front of each of the terms on the left and right like:\n\nx Pb  +  y H2O  +  z O2  -->  a Pb(OH)2\n\nSo now we'll write down equations to make sure that we have the same numbers of both sides of each *element*.  I'll do this by writing down the variable in front of each compound in your equation (so x, y z and a) and then multiplying those by the number of times an element exists in that compound.  (So when I'm writing the equation for hydrogen, the H20 term will be 2*y, because y is the coefficient in front of H20 and H occurs twice in that compound).\n\nRemember, the number of any element on the reactants (left) side equals the number of that element on the products (right) side.\n\nFor Pb: x = a, because Pb shows up only once on each side with a subscript of 1, those two compounds will have to be present in the same molar amounts.\n\nFor Hydrogen: 2y = 2a\n\nFor Oxygen: y + 2*z = 2 * a\n\nSo now we have three equations:\n\nx = y\ny = a\ny + 2z = 2a\n\nBut four unknowns, which is fine because we'll just solve for them in terms of one of the variables and then simply choose the last variable to make all the numbers nice.  So solving the equations in terms of a:\n\nx = a\ny = a\nz = (2a - y)/2 = (2a - a)/2 = a/2\n\nSo now we have x, y and z in terms of a free parameter, a.  We'll just choose a = 2, for example, to make x, y and z all turn out as integers.  So now we've chosen:\n\nx = 2\ny = 2\nz = 1\na = 2\n\nSo the balanced equation is:\n\n2 Pb + 2 H2O + O2 --> 2 Pb(OH)2\n\nMake sense?
728397
You need to first balance the equation.\n\n2CO + O2 --> 2CO2\n\nUsing this you find the answer to be D.
853922
These are the right ones:\n\n2K(s) + 2H2O(l) ---> 2KOH(aq) + H2(g)\nH2SO4(aq) + 2KOH(aq) ---> K2SO4(aq) + 2H2O(l)\nCa(OH)2(aq) + 2HCl(aq) ---> CaCl2(aq) + 2H2O(l)
876161
You're right - as written it's impossible to balance.  But if you write\n\na(Ag2CO3) --> b(Ag2O) + c(CO2) + d(O2) you can balance it with the values \na=1, b=1, c=1, d=0.    Meaning that the reaction doesn't generate any oxygen.  The carbonate falls apart into the oxide and gaseous CO2.
727004
(NH4)2Cr2O7 -> Cr2O3 + 4*(H2O) + N2\n\n(NH4)2SO4 + 2*(NaOH) -> 2*(NH3) + 2*(H2O) + Na2SO4\n\nFe2(SO4)3 + 6*(NH3) + 6*(H2O) -> 2*(Fe(OH)3) + 3*((NH4)2SO4)
73319
the earlier equation is not balanced.\nCaO + 2HCl --------> CaCl2 + H2O
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)
701763
First, notice you have two S atoms on right hand side.\nTo get two S atoms on left, you must have two H2SO4 molecules.\nYour equation is now Cu + 2H2SO4 ---> CuSO4 + SO2 + ??\nIf you subtract right from left, you are left with 4H + 2O atoms, for 2H2O\nSo final eq is Cu + 2H2SO4 ---> CuSO4 + SO2 + 2H2O\n\nHope this helps you to approach these equation-balancing problems methodically.
683249
Oh, come on, it doesn't get any easier than this!\n\nSnO2 + 2H2 --> Sn + 2H2O\n\nAbove is balanced. 1 Sn on both sides, 2 O on both sides and 4 H on both sides.\nYou need to work a little harder on these yourself.
869137
2H3PO4 + 3Ba(OH)2 ---> 6H2O + Ba (PO4)2 +  2Ba
813911
HBr ---> H(+) + Br(-)\nH2SO3 ---> H(+) + HSO3(-)\nCH3-COOH ---> H(+) + CH3-COO(-)
900316
This is a common redox question easily balanced by the half-reaction method.   The half-reaction method is taught using a few different styles, so you'd best learn the style that your teacher or professor expects you to learn.  \n\nFirst, identify your half reactions and balance them:\n\n(Cr2O7)2- contains 2 Cr6+ ions being reduced to 2Cr3+, gaining  6 electrons:\n\n6e-  +  2Cr6+  --->  2Cr3+    (net charge now adds up to +6 on both sides)\n\nMn2+ is oxidized to Mn4+ (as MnO2) losing 2 electrons, or\n\nMn2+  ---->  MnO2  + 2e-\n\nNet electron transfer must equal zero, so we must triple the oxidizing reaction to yield 6 e- to balance the 6 e- gained, so e- gained now equals e- lost (that is, 6 e' in, 6 out)\n\nThat gives us the clue to multiply the manganese term by 3:\n\n(Cr2O7)2- + H+ + 3Mn2+ --->  2Cr3+ + 3MnO2  + H2O\n\nInspection shows the Cr, Mn, and O terms to be balanced but the left side of the equation to be down by 1 H+, so let's put a 2 by the H+....\n\n(Cr2O7)2- + 2H+ + 3Mn2+ --->  2Cr3+ + 3MnO2  + H2O\n\nWe think we're finished here, so let's make the important two final checks:\n1.  Mass Balance and 2.  Charge Balance\n\nEach side of the reaction has 2Cr, 3Mn, 7O, and 2H\n(Mass balance OK)\n\nNow add up the net charges on both sides:\n-2  +2  +6 = +6 on the left side...\n+6 + 0 + 0 = +6 on the right.  (Charge Balance OK)\n\nReaction is balanced.\n\nAgain, WARNING:  Some methods teach you to balance in ACID or in BASE.  This problem, by virtue of the H+ presence is in acid.  Balancing a redox equation in base has a couple more steps attached.  Again, consult your text or your instructor as to their preferred style.
691513
Check that both sides of equation match. Here is an example:\n\n3 CH3CH2OH + 2 Cr2O7 2- + 16 H+ --> 3 CH3COOH + 4 Cr3+ + 11 H2O\n\nThere should be an equal amount of H, C, O, Cr in both sides of the equation. \nIt is sometimes difficult to know where to start balancing an equation. In such case you should try to find an element/molecule that does not appear in every molecule on the other side of the equation. Eg. when balancing this equation above, I first balance the number of Cr because it is on its own on the right side. Sometimes you just need to try and play with numbers for a while to find the right answer. Hope this helped...
290031
Balancing chemical equations is easy, but it's a bit of a trial and error process. Basically, you've got the the arrow in the middle, so the number of each element on one side of the arrow has to equal the number of that particluar element on the other side of the arrow. So start off by balancing any unique elements that only appear once on a particular side. Then balance your nitrogens, oxygens and carbons in that order. The last thing that should be balanced is your hydrogens. I can go into further detail if you need me to, I just don't know how much you know. \n\nFor your other question and yield is how much of something you get from a reaction. For instance, if I mix 100 moles of one substance and 100 moles of another substance, I can calculate a theoretical and experimental yield. For instance, I can say that I expect all 200 moles to react and form a single product. Since i expect it all to react, my THEORETICAL yield is 200 moles or 100%. Say I actually run the experiment and I discover that only 100 moles of my intended product formed, then my EXPERIMENTAL yield is 100 moles or 50%.
781970
2Al + 3F2 ----> 2AlF3\nAl + 3OH- ----> Al(OH)3