(a) Al(s) + HCl(aq)  AlCl3(aq) + H2(g)\nLimiting reactant\nHCl\nAl \nAlCl3\n g \nH2\n g\n\n(b) NaOH(aq) + CO2(g)  Na2CO3(aq)+ H2O(l)\nLimiting reactant\nNaOH\nCO2 \nNa2CO3\n g\nH2O\n g\n\n(c) Pb(NO3)2(aq) + HCl(aq)  PbCl2(s) + HNO3(aq)\nLimiting reactant\nHCl\nPb(NO3)2 \nPbCl2\n g\nHNO3\n g\n\n(d) K(s) + I2(s)  KI(s)\nLimiting reactant\nI2\nK \nKI\n g
200028
Try it yourself, its not that bad. \nHere is how:\n1) Set up a stoichometry ratio.\n2) Using the mass that they give you, calculate the theoretical yield of each product.\n3) Whichever product has a smaller theoretical yield, analyze the reactants which caused it to be such.\n4) The one that 'limits' the amount of products is the limiting reagent.\n5) Using that value, calcuate the amount of the other reactant that is necessary. Subtract from the orginal value to find the excess reactant.
540046
1 - NH4OH\n2 - PbO\n3 - KBr\n4 - PCl5 (five)\n5 - Al(NO3)3\n6 - Ba3(PO4)2  (not sure)\n7 - C4H10\n8 - Sn(OH)4\n9 - NH4NO3\n10 - Li3P\n11 - Ba3ClO2   (not sure)\n12 - MgSO3
385537
NH3OH\nPbO\nKBr\nPCl4\nAl(NO3)3\nBa3(PO4)2\nSpelling error?? C4H10 ??? (not sure here)\nSn(OH)4\nNH4NO3\nLi3P
213781
I suspect that the product is AlSO4.\n\nI'm thinking that the Al swaps with the Cu.  That would leave a free Cu+ which might displace the H from H20 and form either CuOH or CuO, either one of which would probably produce H2.  \n\nSo, I guess you would first get CuOH and AlOH, as intermediate reactions.\n\nI think the equation is:\n\n                          NaCl (goes over the arrow)\n6CuSO4 + 4Al + 12H2O ---------------> 2Al2(SO4)3 + 6Cu(OH)2 + 6H2
391420
CO2\nCarbon Dioxide\nHere is the reaction:\nHC2H3O2(aq) + NaHCO3(aq) ----> NaC2H3O2(aq) + H2O(l) + CO2(g)
403066
2Al + 3F2 ----> 2AlF3\nAl + 3OH- ----> Al(OH)3
269930
a.  4H20 + 4NO2 + 02--------->4HNO3 + 2 H2O\n      Al(OH)3+ 3HCl------------> AlCl3+ 3H2O\nb. not sufficient as Al(OH)3 is 12/78 moles which is able to react with .462 moles of HCl and here we have .9444 moles of HCl\n\n9. it is done by equilliberium reaction\n and u written Na reaction with H2 to form hydride
183108
1) H3COCH3 (or C2H6O) + 3O2 (g) --> 2CO2 (g) + 3H2O (g)\n2) HNO2 (aq) + NaOH (aq) <--> NaNO2 (aq) + H2O (l)\n3) 2K+ + Cr2O7(2-) + Fe(NO3)2 + 2H+ <--> CrO4(-) + Fe3+ + 2K+ + 2NO3(-) \n4) Na3PO4 + NH4NO3 = No reaction, all soluble salts.\n5) 2HCl (aq) + Na2S (s) --> 2NaCl (aq) + H2S (g)
399229
You need to write first the half reactions:\n\nAl ---> Al(+3)\nO2 ---> 2O(=)\n\nAdd the electrons (this is the answer):\n\nAl ---> Al(+3) + 3e(-)\n4e(-) + O2 ---> 2O(=) \n\n{let's check}\nMultiply the first equation by 4 and the second by 3 and sum to eliminate the electrons:\n\n4Al + 3O2 ---> 4Al(+3) + 6O(=)\n\nRecombining the Al(+3) and the O(=) you get:\n\n4Al + 3O2 ---> 2Al2O3.
462271
Li2O + CO2 ---> LiOH\nNa2O2 + CO2 ----> Na2CO3 + O2\nKO2 + CO2 ---->  K2CO3  + O2
366870
1 Aluminum (III) Oxide\n\n2 Iron (II) Sulfide\n\n3 Silver (I) Nitrate\n\n4 Tin (IV) Nitrate\n\n5 Carbon Monoxide\n\n6 Magnesium Sulfide\n\n7 Lead (II) Oxide\n\n9(8?) Sulfur Hexafluoride\n\n10 Mercury (I) Chloride
531787
HCl_aq + NaOH_aq -> NaCl_aq + H2O_l\n(H+)_aq + (Cl-)_aq + (Na+)_aq + (OH-)_aq -> (Na+)_aq + (Cl-)_aq + H2O_l\n(H+)_aq + (OH-)_aq -> H2O_l\n\nFor all the other equations, the (Cl-), (K+), (NO3-), (Na+) are spectator ions, and the n.i.e is the same:\n(H+)_aq + (OH-)_aq -> H2O_l
238553
a) Toluene and Cl2 form p-chlorotoluene and HCl in presence of AlCl3\nb) nitrobenzene + Cl2 -(AlCl3)-> m-chloro-nitrobenzene\nc) chlorobenzene + Cl2 -(AlCl3)-> p-dichlorobenzene\nd) 1-(3-chlorophenyl)ethanone\ne) 3-chlorophenyl acetate\nf) methyl 3-chlorobenzoate\n\nOrganic Chem rocks!
544299
1) Fe(s) +Cu^2+(aq) --> Fe^2+(aq) + Cu(s)  Single replacement, redox reaction where iron is oxidized and copper is reduced\n\n2) 2Al(s)+(3/2)O2(g)-->Al2O3(s)\nredox reaction, oxygen is reduced (neutral to -2) and aluminum is oxidized (neutral to +3)
440498
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)
444434
It can either be a strong acid, like HCl, or a strong base, like NaOH (aq.)  The reactions are:\n2Al + 6HCl => 2AlCl3 + 3H2\n2Al + 2NaOH + 2H2O =>2NaAlO2 + 3H2
372647
(a): Si+, P2+\n\n(b) Ar+, S-\n\n(c) Ca+, Cl2-\n\nBut the other answers that you gave are also correct.