In the combustion analysis of 0.1127 g of glucose (C6H12O6), what mass, in grams, of CO2 would be produced?
958386
C6H12O6 + O2 ----> 6CO2 + 6H20\n\nn(C6H12O6)= 0.1127 / 180 = m / Mw = 6.3x10^-4\n\nglucose to carbon dioxide ratio is 1 to 6 \n\nso 6 x (6.3x10^-4) = n(CO2) = 0.00375\n\nm(CO2) = n(CO2) x Mw(CO2) = 0.00375 x 44 = 0.165g \n\nC
869561
C57H110O6(l)+109O2(g)--heat>\n57CO2(g)+55H2O(l)\n\n\nabout  15.05kg of  CO2
759497
1) determine the mass of carbon, hydrogen and oxygen.\ncarbon: 7.432 g x (12 / 44) = 2.027 g\nhydrogen: 4.057 g x (2 / 18) = 0.451 g\noxygen: 3.383 - 2.027 - .451 = 0.905 g\n\n2) convert to moles.\ncarbon: 2.027 / 12 = 0.169 mol\nhydrogen: 0.451 / 1 = 0.451 mol\noxygen: 0.905 / 16 = 0.0566 mol\n\n3) divide each molar amount by the lowest value to obtain small integers\ncarbon: 0.169 / 0.0566 = 3\nhydrogen: 0.451  / 0.0566 = 8\noxygen: 0.0566 / 0.0566 = 1\n\nThe empirical formula of the substance is (C3H8O)n\nThe atomic masses add up to 60, which is also the molar mass of the compound. Therefore n=1 and we find C3H8O
815144
64 grams... multiply 88 grams by the ratio of molar masses of O2 to CO2 to get  the result
843441
The answer is c.\nC12H22O11+12O2-->12CO2+11H2O
1166564
CO2+H20+sunlight=\nO2+C6H12O6(SUGAR)\nraw material
897570
Dimensional analysis is easier than it first seems.\n\nFirst you need to know the molecular weight of CO2.\n\nCarbon +2(Oxygen) = 12 + 2(16) = 44 grams per mole\n\n[The above step is not dimensional analysis.]\n\nLet's say somebody asks you how many moles 440 grams of CO2 is.\n\nThis is where we do dimensional analysis.\n\n(1 mole CO2 / 44 grams CO2) * (440 grams CO2 / ? CO2)\n\nThe / sign indicates "divided by", so imagine the parts over each other.  That makes it easier but I cannot do it here in text.\n\nThe first part of the above equation is stating that for every mole of CO2, there are 44 grams of CO2.  The second part says that we have 440 grams of CO2, and want to find out how many moles we have.\n\nThe trick here is to have the terms, one above the other, so that the units cancel out and you end up with the desired unit.  Our desired unit here is the mole - we want to know how many moles are in 440 grams of CO2.  So, that is why in the above, (1 mole CO2) is on top of (44 grams CO2).  If we allow 440 grams to be on top in the next term, then grams cancels out, and moles is the only unit left.  This is what we want.\n\nSo if we solve the above\n\n(1/44)*(440/1) = 440/44 = 10 moles of CO2.\n\nSo, 440 grams of CO2 is the same as 10 moles of CO2.\n\nhope this helps\n\nyay
986204
For number 3 its just basic stoichiometry....\nThe molar mass of C6H14 is 86.0g\n\n2.50 g C6H14 (1mol / 86.0 g ) = .0291 mol C6H14\n\nThe -4163 kJ is for one mole of the C6H14, so you just multiply the two to get the answer...\n\n.0291 mol * -4163 kJ = -121. kJ\n\nSo the enthalpy of combustion for 2.50 g of C6H14 is -121 kJ
1174755
One mole of glucose produces approximately 7.6 moles of ATP and one mole of ATP releases approximately 25 calories.
842808
1) determine the mass of carbon, hydrogen and oxygen.\ncarbon: 9.847 g x (12 / 44) = 2.686 g\nhydrogen: 5.038 g x (2 / 18) = 0.560 g\noxygen: 4.146 - 2.686 - .560 = 0.900 g\n\n2) convert to moles.\ncarbon: 2.686 / 12 = 0.224 mol\nhydrogen: 0.560 / 1 = 0.560 mol\noxygen: 0.900 / 16 = 0.056 mol\n\n3) divide each molar amount by the lowest value\ncarbon: 0.224 / 0.056 = 4\nhydrogen: 0.560 / 0.056 = 10\noxygen: 0.056 / 0.056 = 1\n\nThe empirical formula of the substance is (C4H10O)n\nThe atomic masses add up to 74, which is also the molar mass of the compound. Therefore n=1 and we find C4H10O
761721
Take the 125g ethanol (E), C2H6O, and multiply by 1 molE/46gE.\nThe gE cancel, leaving moles E. For the CO2 part, multiply by 2 mol CO2/1 molE. That comes from the balanced equation. The molE cancel, leaving molCO2. Next multiply by 44gCO2/1mol CO2. The molCO2 cancel, leaving gCO2. That's the first part of the problem. Next, go back to the beginning where you got molE. Multiply by 3molH2O/1molE. This also comes from the balanced equation.The molE cancel, leaving molH2O. Multiply by 18gH2O/1molH2O. The molH2O cancel, leaving the gH2O, which is the second part of the problem. At. wts. C=12, H=1, O=16, C2H6O=46
810837
Just remember glucose (C6H12O6). Then using the basics of balancing any chemical equation you can figure out the rest of it.
914456
okay...100 g*(mol/12.01g)=8.33 mol C\n\ntake the aforementioned 1mol *(-113kj/2mol)=-56.5 kJ released for 1 mol\n\nthen multiply that -56.5 by 8.33= -470.65 kJ released\nnote it is exothermic and thus negative
54895
To be precise, the answer is no. Do u noe why? The mean mol wt of air is 29 whereas that of CO is 28.\n\nlet me show u how i get these answers:\nC=12,O=16,mol wt of O2=32g/gmol,mol wt of N2=28g/gmol\nmol wt of CO = 12+16\n             =28\nas for air, by common sense, we know that air is made up of O2,N2, and other gases,rite? As the mol composition of other gases are very negligible(<1%), we assume air is only made up of O2 and N2. Air consists of 21mol% O2 and 79mol% N2.\n\nlet's take a basis of 100gmol(or 100mol)\nmoles of O2=21gmol\nmoles of N2=79gmol\nmass of O2=21X 32g/gmol\n          =672g\nmass of N2=79X28g/gmol\n          =2212g\ntotal mass= 2212+672\n          = 2884g\naverage mol wt of air=total mass/total moles\n                     =2884g/100gmol\n                     =28.84g/gmol\n                     =29g/gmol(whole no.)\n\nsince the ave mol wt of air is greater than that of CO, so aCO is not heavier than air,instead,slightly lighter than air.\n\nDoes it answer ur question?
752459
This is a complete combustion.\nReaction:            aX + bO2 --> dCO2 + eH2O\nd = 0.811818182, derived from mass of CO2 and its Mr.\n\nX has 0.811818182 mol of C.\n\nOriginal sample has 0.811818182/ 1.626 x 60.095\n                                  = 30.00382143mol of C
550253
This time we use oxygen to burn the sugar:\n\nSugar + Oxygen = Carbon Dioxide + Water + Energy\n\nUsing glucose:\nC_6H_12O_6 + 6 O_2 -> 6 CO_2 + 6H_2O + Energy
697475
1) determine the mass of carbon, hydrogen and oxygen.\ncarbon: 7.849 g x (12 / 44) = 2.141 g\nhydrogen: 3.213 g x (2 / 18) = 0.357 g\noxygen: 3.215 - 2.141 - .357 = 0.717 g\n\n2) convert to moles.\ncarbon: 2.141 / 12 = 0.178 mol\nhydrogen: 0.357 / 1 = 0.357 mol\noxygen: 0.717 / 16 = 0.0448 mol\n\n3) divide each molar amount by the lowest value to obtain small integers\ncarbon: 0.178 / 0.0448 = 4\nhydrogen: 0.357  / 0.0448 = 8\noxygen: 0.0448 / 0.0448 = 1\n\nThe empirical formula of the substance is (C4H8O)n\nThe atomic masses add up to 72, which is also the molar mass of the compound. Therefore n=1 and we find C4H8O