how to balance Sn(NO3)4 + AlCl3?
105132
3 Sn(NO3)4 + 4 AlCl3 --> 4 Al(NO3)3 + 3 SnCl4
944650
Are you in chemistry?  Because I am. \nI think it's  3 Al2Cl + 3 Ni. \n\n I hate Cemistry.
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
670034
Ag+ will precipitate in HCL solution as AgCl, but as for Ag(NO3), i am not so sure.
910345
The formula would be Al2(CO3)3; the numbers are subscripts. \n\nLearn your charges for the polyatomic ions and for the other elements, it will really help.
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.
825401
First of all I assume  Na2Co3 should be Na2CO3.\n\nFe(CO)5 + 4NaOH = Na2Fe(CO)4 + Na2CO3+ 2H2O\n\nFirst, balance out the metals (Na and Fe) then C, which was already balanced. Then I chose to balance out H and which automatically balanced out O.
903130
Na2SO4(aq)+ 2 AgNO3(aq)-->Ag2SO4(s)+ 2 NaNO3(aq).
843792
Be very careful:\nFirst, look at the electronegativity differences, the presence of groups like OH- or cations (like Na+ of soap)\nThen, you must know molecular geometries, it is highly important:\nFor BF3 this is a  trigonal planar compound, all differences of electronegativity B<F are canceled due to the position of the F atoms (every 120º on the same plane). It is not polar.\n[SnCl3]- this is a trigonal pyramid compound.\nCl\n I\n:Sn-Cl\n I\nCl\nAS you can see, Sn has a lone pair of electrons, making Cl atoms accommodate in the base corners of an imaginary pyramid. Even if Cl cancel each others dipolar momentum on the sides, they all point ↓. This is a polar compound.
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.
39171
2Al + 3S --> Al2S3\n\nConvert the grams to moles and use stoichiometry to get the moles of Al2S3.
832296
A) Ag2O (s) >>> 2 Ag(s) + 1/2 O2 (g)\nB) Nickel (II) Carbonate >>>>>Nickel (II) Oxide + [Carbon Dioxide], NiCO3 -> NiO + CO2\nC) Ammonium Nitrate >>>>> Dinitrogen monoxide + Water,  \nC does not balance as is... but this does:\nNH3NO3 -> N2O +  1.5 H2O + 0.5 O2\n\n48) A) 2 HCl (aq) + Ca (OH)2 (aq) >>>> 2 H2O + CaCl2 (aq)\nB) 3 Ag2 SO4 (aq) + 2 AlCl3 (aq) >>>> 6 AgCl (s) + Al2(SO4)3 (aq) (silver chloride is a precipitate)\nC) H2C2O4 (aq) + 2 KOH (aq)>>>> 2 H2O + K2C2O4 (aq)\nD) CdBr2 (aq) + Na2S (Aq)>>> CdS(s) + 2 NaBr (aq) (Cadium sulfide is a precipitate)\n50 Write a Balanced equation for the complete combustion of each compound:\nA) Butene (C4H8): C4H8 + 6 O2 -> 4 CO2 + 4 H2O\nB) Octane (C8H18): C8H18 + 12.5 O2 -> 8 CO2 + 9 H2O\nC) glycerol (C3H8O3): C3H8O3 + 3.5 O2 -> 3 CO2 + 4 H2O\nD) Acetone (C3H6O): C3H6O + 4 O2 -> 3 CO2 + 3 H2O\n52: Complete these:\nA) Al (s) + H2SO4 (aq)>>>> AlSO4(aq) + H2(g)\nB) 2 HCl (aq) + Ba (OH)2 (aq)>>>> BaCl2 (aq) + 2 H2O\nC) Au (s) + 3 HCl (aq) >>>>> AuCl3(aq) + 1.5 H2 (g)\n54) EACH IS BALANCED BUT "INCORRECT", so correct these:\nA) Cl-2 + NaI(aq) >>> NaCl2(aq) + I-\nB) NH3(aq) >>> N- + H+3\nC) Na (s) + O2(g) >>NaO2(s)\nD) 2Mg (s) + H2SO4(aq) >>Mg2SO4(aq) + H2(g)\nE) MgCl2(aq) + Ca(OH)2(s) >>> Mg(OH)2(s) + CaCl2(aq)\nF) H2 (g) + Cl2 (g) >> H2Cl2 (g)
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.
820519
First write the balanced chemical equation:\n\nPb(NO3)2 + 2NaBr --->  PbBr2  +  2NaNO3 \n\nThen, find the total volume of the solution: 200mL + 200mL = 400mL \n\nFrom the density, the mass is:   density = mass/volume   mass = density * volume = 1.0 g/mL * 400 mL = 400 g\n\nQ = mC(Tf - Ti) ; where Q = heat released, m = mass of solution,, C = specific heat capacity, and (Tf - Ti) = change in temperature\n\nSo,  Q = 400g * 4.2 J/gK * (2.44 + 273)  (Since C is J/gKelvin, the temperature rise also has to be in Kelvin)\n\nQ = 462.5 KJoules\n\nHope this helps...
874480
2AgNO3 + CaCl2 --> 2AgCl + Ca(NO3)2\n\nMoles of AgNO3 given\n= (115/1000) dm3 x (0.23) mol/dm3 = 0.02645 mol\n\nMoles of CaCl2 given\n= (115/1000) dm3 x (0.14) mol/dm3 = 0.0161 mol\n\n2 moles of AgNO3 needs 1 mole of CaCl2 for complete precipitation. Hence for 0.02645 mol of AgNO3  provided, (0.02645/2 = 0.013225) mole of CaCl2 is needed.\n\n2 moles of AgNO3 produced 2 moles of AgCl. Hence 0.02645 mole of AgCl is produced.\nMass of 0.02645 mole of AgCl\n= (0.02645 x Mr of AgCl) = 0.02645 x 143.5 = 3.80g (to 3 s.f.)\n\n0.02645 mole of AgNO3 produces 0.02645 mole of Ag+ and 0.02645 mole of nitrate ions.\n0.0161 mole of CaCl2 produces 0.0161 mole of Ca2+ and (0.0161 x 2 = 0.0322) moles of Cl-.\n\nAfter precipitation, with AgNO3 as the limiting reagent,\nMoles of Ag+ left = 0 (all precipitated as AgCl)\nMoles of Cl- left = 0.0322 - 0.02645 = 0.00575\nMoles of nitrate ions left = 0.02645 (nitrate not precipitated)\nMoles of Ca2+ = 0.0161 (Ca2+ not precipitated either)\n\nTotal volume of solution is now (115 + 115 = 230 ml).\nFor concentration of each ion left, please divide the moles of each ion left by the volume (230 ml) respectively.\n\nHope this helps. :)
852560
Sn has a valence (charge) of +3, that is also its' oxidation number.\nSn had 3 valence electrons it lost so the next lowest level would have eight. Having lost the three electrons gave it a +3 charge.  \nNormally Sn has +4 or +2 as a oxidation number. It's charge in this molecule is +3 (an odd occurrence), I think it is written incorrectly, having never seen Sn as +3.\n\nCl has a charge of -1, it gained one electrons because it had 7 valence electrons, gaining the one gave it eight. \n\nA compound must be neutral, so three Cl are needed to counter the one Sn.
889156
The answer is BrF3.\n\nand the balance equation is \n\nBr2  +  3F2  --->  2BrF3\n\nhow? get the moles of each using this formula:\nn = PV/RT\n\nyou'll get:\nmol Br2 = 0.041mol\nmol F2 = 0.123 mol\nmol pdt = 0.082 mol\n\nbase from this, you'll found that ratio of Br:F is 1:3. Hence, the product is BrF3. you'll also found that ratio of Br2:F2:pdt is 1:3:2 which also agrees with our balance equation above.