Balance this equation Rb(s)+Cl2(g)--RbCl(s)?
956142
2Rb(s)+Cl2(g)--2RbCl(s)
923888
CaCO3(s) + 2 HCl(aq) --> CaCl2(aq) + CO2(g) + H2O(l)
998581
The answer is A\n\nCoBr2 + Pb(NO3)2 ==> PbBr2  + Co(NO3)2\nPb(C2H3O2)2 + 2KBr ===> PbBr2 + 2KC2H3O2\n\nIn both cases, the net ionic equation is:\n\nPb+2  + 2 Br-  ====> PbBr2(s)
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...
160963
K2SO4 ---> 2K+  +  SO4-\n\nHCl ---> H+  +  Cl-\n\nNaHCO3 ---> Na+  +  HCO3-\n\nNH4OH ---> NH4+  +  OH-\n\nBaCl2 ---> Ba-2  +  2CL-\n\n(NH4)3PO4 ---> 3NH4+  +  PO4-3
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.
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.
813421
Na2CO3(aq) + MgCl2(aq) --> 2NaCl(aq) + MgCO3(aq)
50281
This is an acid base reaction, but the other answer is incorrect about the product. There is no Cl present. Here’s the balanced equation:\n\n2NaOH (s) + H2CO3 (aq) ➔ Na2CO3 + 2H2O\n\nNaOH is the base. H2CO3 is the acid.
743385
Cl2 + 2 KI → I2 + 2 KCl
739042
2Al + 6HNO3 ----> 2Al(NO3)3(s) + 3H2(g)
564798
Ba(NO3)2 + Na2CO3-->BaCO3 +2NaNO3\n\nto do a double displacement follow this formula\n\nAB + CD --> AD + CB
382458
H2 + Cl2 -> 2HCl\n2Ag + H2S -> Ag2S + H2 \nCH4 + 2O2 -> CO2 + 2H2O\n2HgO -> 2Hg + O2\n3Fe + 4H2O -> Fe3O4 + 4H2
755565
1.) Cr(NO2)2+(NH4)2SO4\n---->CrSO4+(2)NH4NO2\n2.) (2)KNO3+H2CO3\n----->K2CO3+(2)HNO3\n3.) (3)KBr+Fe(OH)3\n----->(3)KOH+FeBr3\n4.) B2Br6+(6)HNO3\n----->(2)B(NO3)3+(6)HBr\n\nHope this helps=)\nTip: Start by balancing those more isolated atoms first, especially when you are balancing more complex equations (they may have atoms that appear in more than 1 products).\nAlso, feel free to leave numbers in fractions while you balance ur equation. You can always multiply all the numbers later to make them all whole numbers. You will find this useful later when you have to balance long equations. Itd definitely more tedious to alter every number in ur mind and writing them all out at one go. Do it progressively=)
1158541
That's easy:\n\n2 AlCl3 + 3 Pb(NO3)2 ===> 2 Al(NO3)3 + 3 PbCl2\n\nMultiply both sides of your equation by two:\n\n4 AlCl3 + 6 Pb(NO3)2 ===> 4 Al(NO3)3 + 6 PbCl2
372026
CuSO4·5H2O(s) → CuSO4(s) + 5H2O(g)
986618
The equation is already balanced. Count the Ag's on each side = 1. Count the Na's on each side = 1. Count the Cl's on each side = 1. Count the NO3's on each side = 1. One can count the N's on each side = 1, or the O's on each side = 3.\n\nBut one could also consider the question another way:\nAg+ + NO3- + Na+ + Cl- ===> AgCl + Na+ + NO3-