balancing equation Cu+HNO3--->CU(NO3)2+H2O+NO2?
606075
Cu(s) + 4HNO3 (aq)--> Cu(NO3)2(aq) + 2H2O(l) +2NO2(g)
844624
FeCl3   +    3NaOH  ----->    Fe(OH)3   +   3NaCl\n\nthats it
372026
CuSO4·5H2O(s) → CuSO4(s) + 5H2O(g)
927756
2H2O2 → 2H20 + 02 is the balanced equation for the above decomposition
775066
You have the products, but now you have to balance the equation. The unbalanced equation is:\n\nLi2CO3 + Cr2(SO4)3 --> Li2SO4 + Cr2(CO3)3\n\nBalanced:\n\n3 Li2CO3 + Cr2(SO4)3 --> 3 Li2SO4 + Cr2(CO3)3
651011
H2O2 + OC(NH2)2 -> HCOOH + NH4OH -> NH4(HCOO)
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?
748872
Cu2+ (aq)  +  Zn (s) -->  Cu (s) + Zn2+ (aq)
945429
Equation becomes CuCl2 + 2KI --> CuI2 + 2KCI\n\nProducts are Cuprous Iodide and Potassium Chloride
116069
The balance is :\n\n     H2O2(aq) -> O2(g) + 2H+(ag) + 2e-\n MnO2(s) + 4H+(aq)  +  2e-  ->  Mn2+(aq)  + 2H2O\n \n Combine the two half-reactions and cancel on both sides yield the     complete balance equation.\n\n MnO2(s) + H2O2(aq) + 2H+(aq) -> Mn2+(aq) + O2(g) + 2H2O\n\n this is an oxidation-reduction reaction\ntype  decomposition reaction.
894420
Here you are:\n\nCu2S + 12HNO3 => Cu(NO3)2 + CuSO4 + 10NO2 + 6H2O
596804
3Cu+8HNO3=3Cu(NO3)2+2NO+4H2O
394775
2Cu(NO3)2 ----> 2CuO + 4NO2 + O2\nOBSERVATION: brown, choking fumes of NO2 evolved; black solid i.e CuO settles at the bottom\n\n2FeSO4 ----> Fe2O3 + SO2 + SO3 \nOBSERVATION: white fumes of SO3 evolved and reddish brown solid i.e. Fe2O3 formed\n\nEDIT: The above equation for iron(II)sulphate may be a supposed characteristic for its thermal decomposition relating to that of copper(II)nitrate above, but Chaos, O2 will not be evolved here, because SO3 will not get reduced to O2, even on strong heating.
929071
2C18H18 + 45 O2 --->  36 CO2 + 18 H2O\n3Cu + 8HNO3 ----> 3Cu(NO3)2 + 2NO + 4H2O
739042
2Al + 6HNO3 ----> 2Al(NO3)3(s) + 3H2(g)
148175
The correct balanced equation is:\n2ZnS + 3O2 --> 2ZnO + 2SO2\nThis is BALANCED\nMr chemistry teacher, you didnt balance your equation,...and i'm in the ninth grade
1097630
Na2CO3 + 2HCl-> NaCl + H2O + CO2