Problems balancing 2H I (aq) + K2SO3 (s) -->?
957626
It is because you have the wrong compound (HSO3)!\n\n2 HI + K2SO3 ---> 2KI + H2SO3
307574
No, it's Ag2S. The 2 is a subscript, denoting 2 Silver atoms to 1 Sulphur atom. Silver has a valency of 1+, while Sulpur has a valency of 2-, hence 2 Silver atoms are needed to balance with 1 Sulphur atom to form a compound. Besides, the equation would make sense then. 2 Silver atoms and 1 Sulphur atom on both sides of the equation, the equation is now balanced. 10 marks please, thank you!
747348
Hi Kitty!  Dr. J to the rescue with a full answer.\n\n1) 4Li + O2 ---> 2Li2O\n\n2) 4HF + SiO2 ---> SiF4 + 2H2O\n\n3) It is simpler if we eliminate the K and the sodium and write the equation ionic because it is a redox reaction and we need to balance.\n\nThe overall redox equation will be:\n\nCr2O7(=) + SO3(=) + H(+) ---> Cr(+3) + SO4(=).\n\nNow you balance by the half reaction method.  The two reactions are:\n\n6e(-) + Cr2O7(=) + 14H(+) ---> 2Cr(+3) + 7H2O\n3SO3(=)  + 3H2O --->  3SO4(=) + 6H(+) + 6e(-)\n\nEliminating the electrons:\n\nCr2O7(=) + 3SO3(=) + 8H(+) ---> 2Cr(+3) + 3SO4(=) + 4H2O\n\nThis one was tough!!!  Thumbs up!\n\nIf you want you could put the K and the Na back.\n\nBest answer?
502184
5H2O2(aq)+2KMnO4(aq)+\n3H2SO4(aq)-->\n2MnSO4(aq)+K2SO4(aq)+\n8H2O(l)+5O2(g)
112894
Haven't had chem in a while but maybe it would help to write out all the individual symbols.  Your second example would look like\n\nHHHPOOOO + MgOOHH  = MgMgMgPPOOOOOOOO + HHO\n\nthen you can see that the left side needs three Mg so you'd need 3 of the Mg(OH)2 to balance.  The left side also needs two P so there needs to be two H3PO4 there.  This makes a total of 12 H on the left, so to balance that you need 6 H2O on the right.  This also gives 14 O on the right to balance the 14 O you'd now have on the left.  So\n2 H3PO4 + 3 Mg(OH)2 = Mg3(PO4)2 + 6 H2O
975769
Al(s) + MnO4- (aq) -----> MnO2 (s) + Al (OH)4- (aq)\n\nSplit the above into two half equations:\n\n1) Al(s)  --->  Al (OH)4- (aq)  and\n2) MnO4- (aq) -----> MnO2 (s)\n\nBalance 1) as follows:\n\n1) 4OH^-  +  Al(s)  --->  Al (OH)4- (aq) now balance for charge:\n1) 4OH^-  +  Al(s)  --->  Al (OH)4- (aq)  +  3e^-\n\nBalance 2) as follows:\n2) MnO4- (aq) ---> MnO2 (s)  +  2H2O (l) (turn xs O into H2O)\nNow balance with H^+:\n2) 4H^+  +  MnO4- (aq) ---> MnO2 (s)  +  2H2O (l)\nnow balance charge:\n2) 3e^-  +  4H^+  +  MnO4- (aq) ---> MnO2 (s)  +  2H2O (l)\n\nNow add 1) and 2) (electrons happen to be equal):\n4OH^-  +  Al(s)  +  4H^+  +  MnO4- (aq) ---> MnO2 (s)  +  2H2O (l)  Al (OH)4- (aq)\nnow OH^-  +  H^+  ---> H2O:\n\n4H2O  +  Al(s)    +  MnO4- (aq) ---> MnO2 (s)  +  2H2O (l)  Al (OH)4- (aq)\n\nSubtract 2H2O from both sides:\n2H2O (l)  +  Al(s)    +  MnO4- (aq) ---> MnO2 (s)  +    Al (OH)4- (aq)\n\nI hope that helps. If you need more help, please email me or re post.
66380
You have to use the formula -\nm1s1(t3-t1)=m2s2(t2-t3)\n=>191(4.184)(30.5-25)=38s2(192-30.5)\n=>s2=191(4.184)(5.5)/((38)(161.5)\n=>s2=0.716 J/g.k
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.
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
764101
In all reactions one uses that 2 OH -> H2O + O, the split of oxygen goes to create the ion. Finally balance the equation:\n \nSO3 + 2 NaOH -> Na2SO4 + H2O\n\nCO2 + 2 KOH -> K2CO3 + H2O \n\nSO2 + Ba(OH)2 -> BaSO3 + H2O
550368
I think you have a notational error: did you mean Al2(SO4)3 for aluminum sulfate?\nIf that is the case, then proceed as follows: \nAl(OH)3 + H2SO4 -> Al2(SO4)3 +H2O\n1. First, balance the Aluminum atoms:\n\n2Al(OH)3 + H2SO4 -> Al2(SO4)3+H2O (2 Al)\n2. Next, balance the Sulfur:\n2AL(OH)3 + 3H2SO4 -> Al2(SO4)3+H2O (3 S)\n3. Next, balance Oxygen:\n2AL(OH)3 + 3H2SO4 -> Al2(SO4)3+6H2O (18 O)\n4. Hydrogen is already balanced (12 H)\nGenerally speaking, the strategy is to balance the least represented element first until you move to the most represented.(here it was aluminum, followed by sulfur). Hydrogen is usually the last one you balance.
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
1097630
Na2CO3 + 2HCl-> NaCl + H2O + CO2
841292
When properly balanced, the coefficients read from left to right, 1,3,2,3. The balancing used by the other two people doesnt even balance correctly!!
904363
(NH4)2CO3+MgSO4->(NH4)2SO4+MgCO3
804535
Neutralization Reaction:\nHC2H3O2(aq) + KOH(aq) -->H2O(l) + KC2H3O2(aq)\nThe equation is already balanced.\n\na. False KC2H3O2(aq) and H2O are products.\nb. True\nc. False, some ions will cancel out but some will remain.\n\nThe ionic equation is as follows:\n(H^+)(aq) + (C2H3O2^-)(aq) + (K^+)(aq) + (OH^-)(aq) -->H2O(l) + (K^+)(aq) + (C2H3O2^-)(aq)\nRepeated ions are crossed off from both sides and the net ionic equation is:\n(H^+)(aq) + (OH^-)(aq) -->H2O(l)\n\nd. True\ne. False\n\nNote: the + and - signs were written with ^ to indicate that they were raised up (like a power).  Sorry if this notation is confusing.  Good luck and I hope that this helps!
967763
Equivalent to [H+]^2/([HCH3CO2] - [H+]).  You can solve for [H+] knowing Ka and [HCH3CO2].