7 C(s) +  9 CH4(g)    --> 2 C8H18 (l)  Is it possible?
913828
It sounds extremely unlikely.  In fact, so unlikely that I would rate it as impossible.  The problem is that the first stage of the reaction would be the splitting of one of the C---H bonds in methane and you would need pretty brutal reaction conditions to achieve this, e.g. temperatures well in excess of 1000°C.  At these temperatures, hydrocarbons such as octane are unstable and simply would not form.  If there is any reaction at all, I would guess that, at fairly high temperatures you would get some acetylene, e.g. by passing a stream of methane over a bed of red-hot carbon :\nC(s)   +   CH4(g)   --->   2 C2H2(g)\n\nNote that the equation you wrote is balanced.  But just because you have a balanced equation, that doesn't mean the reaction is feasible.
48368
2HC2H3O2 + Ba(OH)2 --> Ba(C2H3O2)2 + 2H2O
741784
(21 g x 52.7 C) + (54.9 g x 31.5 C) / (21 g + 54.9 g)\n\n= 37.4 C
262916
4 OH-(aq) --->  O2(g) + 2H2O(l) + 4e-
887797
It looks like you got it. This is how it came out for me:\n\nThis is the reaction:\nC10H8(s) + 12 O2(g) ® 10 CO2(g) + 4 H2O(l)\n\nand Hc=(Summation of Hf of products)-(Summation of Hf of reactants)\nSo...\n\nHc (C10H8) = 10 Hf(CO2) + 4 Hf(H2O) – Hf(C10H8)\n-5156.8kJ/mol=(10)(-393.5 kJ/mol)+(4)(-285.9kJ/mol0-Hf (C10H8)\n-5156.8kJ/mol=(-3935 kJ/mol)+(-1143.6 kJ/mol)- Hf (C10H8)\n-5156.8kJ/mol=(-5078.6)- Hf(C10H8)\n-78.2kJ/mol=-Hf (C10H8)\n78.2kJ/mol=Hf (C10H8)\n\nI believe that's how it's done, but it's been a while since I've done that kind of problem, so it's probably not a terrible idea to double check with someone else.
462663
C4H9OH + 6O2 --> 4CO2 + 5H2O
229945
-> 6CO2 + 6H2O     Carbon dioxide.
954811
C3H8 + 5O2 --> 3CO2 + 4H2O
904363
(NH4)2CO3+MgSO4->(NH4)2SO4+MgCO3
929071
2C18H18 + 45 O2 --->  36 CO2 + 18 H2O\n3Cu + 8HNO3 ----> 3Cu(NO3)2 + 2NO + 4H2O
917667
Part A:\nThe answer is 12 L H2.  Here is the solution:\n(4 L N2)(1 mol N2 / 22.4 L N2)(3 mol H2 / 1 mol N2)(22.4 L H2 / 1 mol H2)\nso (4*1*3*22.4) / (22.4*1*1) = 12 L H2\n**note that in in ideal gas there are 22.4 liters/mol**\n\nPart B:\nThe answer is 8 L NH3.  Here is the solution:\n(4 L N2)(1 mol N2 / 22.4 L N2)(2 mol NH2 / 1 mol N2)(22.4 L NH3 / 1 mol NH3)\nso (4*1*2*22.4) / (22.4*1*1) = 8 L NH3\n\nHope this helps!
289562
First of all, it's spelled stoichiometry.\nSTP=22.4L per mole\nmethane = CH4\nmolar mass of methane = 4*1.0079(mass of 4H) + 12.011(mass of C)=16.0426gCH4\n\nSo it would be like this\n3.35L / 22.4L * 16.042g = 2.39923g methane\n\nWith significant figures, it would be 2.40g methane
818293
im pretty sure but  not positive\nC3H8 + 5 O2  -->  4 H2O + 3 CO2\n\nso 5 mol O2 to 1mol C3H8
775000
2 MnO4-(aq ) + 5 H2C2O4(aq ) + 6 H3O+(aq ) --> 2Mn2+(aq ) + 10 CO2(aq ) + 14 H2O
222097
OH-(aq) + HC2H3O2(aq)---> H2O(l) + C2H3O2-(aq)
122744
Mg + 2 CH3COOH = Mg(CH3COO)2 + H2
731348
C3H8 is CH3-CH2-CH3\n\nthere is no way that this will fragment into CH4.\n\nyou will get fragments of\nCH3 (m/e = 15)\nCH2CH3 (m/e = 29)\n\ncan you see any way to split the above molecule into CH4?\n\nremember that fragmenting means destroying a bond, thus if you were to get CH4, you'd have to have started with CH4 bonded to something, but that breaks the octet rule...