Please answer this chem problem for me?
917127
Here's the how you do it.\n\nBalance the equation.  Element by element.\n\nThat gives you the mole ratios of  the various compounds.\n\nConvert the moles to mass using the molecular weight of each of the compounds.\n\nThat gives you the mass ratios of the various items.\n\nDivide both sides of the equation by the largest mass on the left side of the equation,    Then  you will have the mass of products per Kg of the at least one of the reactants.  Remember  you are starting with 1.0 kg of one item and 0.98 of the other.\n\nso you must check to see which reactant will limit the production.\n\nShould be a simple matter of working out the ratios.
857176
S(s) + O2(g)------>SO2\n\n2Mg(s) + O2(g) -------> 2MgO(s)\n\nH2(g) + Br2 (g) -------> 2HBr(g)\n\n4Fe(s) + 3O2 -------> 2Fe2O3 (s)\n\nThen there are single replacement \n\nA + BC-------> AC + B\nZn(s) + 2HCl(aq)------->ZnCl2 (aq) + H2 (g)\nwhere one element replaces another\n\nthen there is double replacement \n\nAB+CD -------> AD + BC\n\nIn this one the positive ions in the reacting compounds switch places\n\nBaCl2 (aq) + Na2So4 (aq) --------.BaSo4 (s) + 2NaCl (aq)
788070
Well its easy dear.\n\n2Li (s) + Cl2 (g) ----> 2 LiCl (s)
808775
Na2C2O4\n Na +1, C +3, O -2
574078
This might help :)\nhttp://www.chemicalforums.com/index.php?board=8;action=display;threadid=3263
205870
CaCO3(s) + 2HC2H3O2(aq)-> Ca(C2H3O2)2(aq) + H2O(l) + CO2(g)
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.
651011
H2O2 + OC(NH2)2 -> HCOOH + NH4OH -> NH4(HCOO)
295951
2 CH3OH + 3 O2 → 2 CO2 + 4 H2O + 726 kJ/mol
832115
N2H4 + H2O -> N2H5+ + OH-\n\nKb = c[N2H5+] * c[OH-] / c[N2H4]\n\nAssuming that c[N2H5+] = c[OH-] and that c[N2H4] >> c[OH-] (so that we can use the starting concentration for c[N2H4]), we get approximately\n\nc[OH-] = sqrt( Kb * c[N2H4])\n\nor logarithmized\n\npOH = 1/2 ( pKb + 10^-(c[N2H4])\n\npH= 14 - pOH\n\n\ncalculator time.
497024
H2S---> 2 H^+ + S^2-\nK(eq) = \n[2H^+]^2 [S^2-]\n ---------------\n     [H2S]\n\nNH4+----> NH3  + H^+\nK(eq) =\n [H^+][NH3]\n       ------------\n         [NH4+]\n\nC6H5COOH----> C6H5COO-  + H+\n\nK(eq) = \n[H^+][C6H5COO-]\n        ---------------\n         [C6H5COOH]
739541
CH4 + 2O2-> CO2 + 2H2O
195778
1.  C(s)+O2--> CO2(g) - 393.5 kj\n2.  H2(g)+1/2 O2(g)--> H20 -285.8 kj\n3.  2C2H2(g)+5O(g)-->4CO2(g)+2H2O(... -2598.8\nCALCULATE the value for the reaction 2C+H2-->C2H2\n\nFirst balance eqn 3 with 5O2 (g) then reverse it so that you have 2C2H2(g) on the RHS. Of course, you will change the sign of the eqn as well. Now halve it so you have one C2H2 instead of 2. Let's call the new eqn number 4\n\nSecond, double the 1st eqn. Let's call it eqn 5\n\nLastly, add eqns 3, 4 and 5:\n\n4 2CO2(g)+H2O --> C2H2(g)+5O2(g) +1299.4 kJmol-1\n5 2C(s)+ 2O2--> 2CO2(g) -787.0 kJmol-1\n3 H2(g)+1/2 O2(g)--> H20 -285.8 kj\n----------------------------------------------\n  2C(s)+ H2(g) --> C2H2(g) + 226.6 kJmol-1
739042
2Al + 6HNO3 ----> 2Al(NO3)3(s) + 3H2(g)
727458
2. 2 Na + 2 H2O --> H2 + 2 NaOH\n3. 2 HNO3 + Ca(OH)2 --> Ca(NO3)2 + H2O\n4. C5H12 +  8 O2 --> 6 H2O + 5 CO2
747676
2C12H22O11 +24O2 ---> 24CO2 +22H2O\na)F\nb)F\nc)T\nd)F\ne)T\nf)T
843441
The answer is c.\nC12H22O11+12O2-->12CO2+11H2O