butane stoichiometry?
102536
First use the ideal gas law to find out how much oxygen gas reacts with the butane.\n\nPV=nRT where P is pressure in Pa, V is volume in meters cubed (m^3), n is amount in moles, T is temperature in Kelvin (K), and R is the ideal gas constant 8.314 J/mol K.\n\nn=PV/RT=[(101325 Pa)(0.015 m^3)]/[(8.314 J/mol K)(298 K)]\nn=0.613 moles O2\n\nNow write the balanced equation for butane combustion.\n\n2 C4H10 + 13 O2 --> 8 CO2 + 10 H2O\n\nNow use stoichiometry to find the moles of butane.\n\n(0.613 mol O2)(2 mol C4H10/13 mol O2)= 0.094 moles C4H10.\n\nNow convert moles C4H10 to grams by multiplying by molar mass. The molar mass for butane is 58 g/mol.\n\nMass butane = (0.094 moles)(58 g/mol) = 5.47 grams butane
764178
Trick question, multiple answers. . .\n\n1. CH3(CH2)3OH\n2. C2H5CH(OH)CH3\n3. (CH3)2CHCH2OH\n4. (CH3)3COH\n5. (C2H5)2O\n6. CH3(CH2)2OCH3\n7. (CH3)2CHOCH3\n\nOk, #4 is tert-butanol.  All of these are structures for C4H10O\n\ngo to http://www.cem.msu.edu/~reusch/VirtualText/intro3.htm and scroll down to the isomers section.  All the diagrams are listed there.
165982
Propane - C3H8 = 44 grams/mole\nButane - C4H10 = 58 grams/mole\n1 cc = .001 liter\ngas constant R = .08206\n\nA)\n\nPV=nRT\n(1 atm)(50x.001L) = n(.08206)(273K)\nn=.002 mol\n\n.002 mol propane x (44 g/mol) = .088 g\n.002 mol butane x (58 g/mol) = .116 g\n\n50% propane by mole\n43% propane by mass\n\nB)\n\n(150 kPA)/(101.325 kPA) = 1.48 atm\nPt = P1 + P2\n1.48 atm total = .74 atm propane + .74 atm butane\n\nC)\n\nThe propane should effuse faster because its molecular mass is lower.
714714
A balanced equation for this combustion is:\n\nC3H8 + 5 O2 ----> 3 CO2 + 4 H2O\n\nSo, every mole of C3H8 requires 5 moles of O2 to burn.\n\n# of moles of C3H8 = 138 / (12*3+8) =3.136 mole\n\n# of moles of O2 required = 3.136*5 = 15.682 moles\n\nUsing ideal gas law, PV = nRT, we get\n\nV = nRT / P = 15.682*8.314*(25+273) / (100*1000)\nV = 0.389 m^3
121351
Are you tryn to have us do your hw!?!?\n\nThe eqn should be in your book.  I dont feel like digging mine out, but its something like;\n\nCp*g*(T2-T1)=energy\n\nSo what u want to find is (t2-t1) Ima call it x\n\n400*50*(x)=800  do the math.  I think thats it.\n\nfor the 2nd question, this one is easy.  The combustion is energy is for 1 mol (per mol).  If you only have .2 moles, the combustion amount is .2* 1370 .
947798
well, its all in the stoichiometry.\nsince u have the balanced equation, just do it out!\nuse 1 gall C8H12 * the equavalent in liters (convert 1 gallon to liters) and then u can do the stoichiometry going from liters of C8H12 to grams of oxygen using the molar ratio. \ngood luck! hope this helps!
221143
They're gases (although butane forms a liquid under a little pressure) and the rest are liquids, at room temperature. Pentane is highly volatile, though.
966248
n-butane\n/\n&#92;\n/\n\ni-butane\n&#92;_\n/\n\nBest I could do here.
939108
Given:\nPV = nRT, 862 torr x (1 bar/750torr) = 1.15 bar, 117 C +273 = 390 K, CH4 = 16 g/mol\n\nn/V = P/(RT)\n\nn/V = (1.15bar)/[(0.08315 L bar/K mol)x(390K)\nn/V = 0.035 mol/L x (16 g/mol) x (1L/1000mL)x(1mL/1cm^3)\ng/cm^3 = 5.67 x 10^-4 g/cm^3
215604
C6H12O6(s) ---H2SO4--> 6C(s) + 6H2O(in H2SO4 solu.) + Heat
1068778
pure butane is neutral for sure. And butane, no matter isobutane(2-methylpropane) or just butane, is an alkane. \n\npH = -log[concentration of hydronium ions] (hydronium is H3O+), but Alkane does not dissolve in water since no ionic bond exists in the compound, so measuring its pH does not make much sense.\n\nCarbon and hydrogen have very similar electronegativity, so it's very difficult to remove hydrogen from the saturate carbon chain. alkane has pka value around 50, pKa value for water is around 16.
686129
C2H2 - acetylene\nC2H6 - ethane\nCH4 - methane\nC8H18 - octane\nC2H3OH - ethanol\nthis list is a mile long
453509
i'm not familiar with the unit barg, but just use the ideal gas equation PV=nRT solved for n so n=(PV)/(RT) with an apropriate value for the universal gas constant (R).  that will give you the number of moles of N2 which you can then determine the volume under standard conditions if you wish using the same equation.
1148552
It can be estimated but without a detailed analysis, an accurate determination can not be made. \n\nNatural gas for instance can contain various contaminates that will lower the heating value and some heavier hydrocarbons that may raise the heating value. \n\nDiesel or gasoline can also have variations depending on the exact composition. Both are multi-component fuels with varying amounts of certain hydrocarbons and as a result the heating value can vary.\n\nLPG (ie. propane, butane or propane/butane mixutures) can vary in composition.  As an example not only is the amount of propane in a propane product different from tank to tank but also the amount of ethane and butane. depending on the degree to which these vary there can be some variation in the heating value. \n\nThe real question is how accurate do you have to be with your answer.
158793
Okay to answer the stoichiometry problem first...You need to balance the equation first...If you don't know how to balance the equation email me and I will explain.  But the balanced equation is:\n\n2Na3PO4 + 3Ba(NO3)2 ----> Ba3(PO4)2 + 6NaNO3\n\nHere is a basic "map" you need to follow:\n\nGrams of Substance A divided by molar mass of A ===> Moles of A times the coefficients of A and B from balanced equation to get moles of substance B===> Moles of substance B times molar mass of substance B ===> Grams of substance B\n\n\nTo start we need to calculate the molar mass of the two compounds we are mixing together\nNa3PO4 = 151.97g/mol\nBa(NO3)2 - 213.34g/mol\nThen we need to find the molar mass of the product, which is Ba3(PO4)2 - 577.97g/mol\nIf you don't know how to find the molar mass, email me. Otherwise, read a chemistry book.  Anyway the steps are to convert grams to moles then multiply by the coefficients of our balanced equation and then multiply again by grams of \n Ba3(PO4)2...Seems complicated, but it's really not.  \n\n1. To see if Na3PO4 is the limiting reactant we divide 3.50g by it's molar mass of 151.97g/mol to get moles of Na3PO4\n\n3.50/151.97 = 2.30*10^-2 mol. Take this and multiply by the coefficients of our reactant and product:\n\nTake the coefficient in front of Na3PO4 and place that in the denominator  /2 \nTake the coefficient in front of our product Ba3(PO4)2 and put it in the numerator  1/\n\nNow take the calculated moles of \n2.30*10^-2 and multiply them by (1/2) <----  the coefficients of reactant and product\n\nThis gives you moles of Ba3(PO4)2\n(2.30*10^-2) * (1/2) = 1.15*10^-2 mol\nNow multiply this number by the mol weight of our product which is 577.97g/mol and that gives you grams of product produced.\n\n(1.15*10^-2) * 577.97 = 6.66g of Ba3(PO4)2\n\nWhew!\n\nNow on to the next one....I won't explain fully as it's exactly the same procedure.\n\n6.40g/213.34g/mol = 2.99*10^-2 mol of Ba(NO3)2 (reactant)\n\n(2.99*10^-2) * (1/3) = 9.99*10^-3 mol of Ba(PO4)2 (product) \nI used (1/3) because the number in front of the reactant is 3 (goes in the denominator) and the 1 is the coefficient in front of the product.\n\nNow   (9.99*10^-3) * (577.97) = 5.78g of Ba3(PO4)2\n\nokay so upon inspection, the limiting reactant is Ba(NO3)2 and it yields 5.78 grams...\n\nYou ask if 3.50 grams are actually yielded, what is the %\n\nSo....(actual yield/theoretical yield)*100 = % \n\n(3.50/5.78)*100 = 60.55%\n\nAlrighty....Now onto problem #2\n\nBalance: Al(NO3)3 + Na2S ----> Al2S3 + NaNO3      \n\n(A) Al(NO3)3 + (B) Na2S ----> (C) Al2S3 + (D) NaNO3\n\nPut variables in front of each reactant and product like I did above.  Now assign an element to equal the variable on both sides.  In our case:\n\nAl  -    A = 2C\nNO3 -   3A = D\nNa  -   2B = D\nS   -   B = 3C\n\nNow make one letter equal one. We'll make A = 1\nSo..\nA = 1\nB = 3/2  (by algebra)\nC = 1/2  (by algebra)  (email me if you don't know)\nD = 3\n\nwe can't have fractions in there, so we multiply all of them by two to get the 2 in the denominators out, and we end up with\n\nA = 2\nB = 3\nC = 1\nD = 6    With a final equation balanced out to look like:\n\n2Al(NO3)3 + 3Na2S ---> Al2S3 + 6NaNO3\n\nHope all this was clear!!!  Enjoy
199454
Gas is octane  is nontane m=9 n=21\nIt may not be so pure.
373250
C3H7NO2 \n\nwww.chemfinder.com is a useful resource.  \n\nI hope this helps :)