show work for the number of grams of oxygen necessary to burn 100 grams of C4H10. 2C4H10+13O2---> 8CO2+10H2O
96968
mass of C4H10 = 48.0amu + 10.1amu = 58.1 amu\n(100.0g C4H10/1)(1 mole C4H10/58.1g)= 1.72 moles C4H10\nSo 100 grams of C4H10 is 1.72 moles.\n\n(1.72 moles C4H10/1)(13 moles O2/2 moles C4H10)= 11.2 moles 02\nSo you need 11.2 moles of oxygen to burn 100 grams of C4H10.\n\n(11.2 moles O2/1)(32.0g/1 mole O2)= 358g O2\n\nYou need 358 grams of oxygen.
823190
This is a stoichiometric problem.  \n\nFirst, find the the amount in grams of one mole of Fe2O3.\n\nFor the first problem, convert the 3.02g Fe2O3 to moles.\n\nThen set up dimensional analysis.  For every 2 mol of Fe2O3, you need 2 mol C.  So you need an equal amount of moles for both.\n\nTake the amount of moles you have of Fe2O3 and convert them to grams Carbon.  \n\nFor the second problem, you should do the exact same thing except you dont need to convert to moles in the beginning because it's already in moles!\n\ni hope this is helpful!!!
843441
The answer is c.\nC12H22O11+12O2-->12CO2+11H2O
407297
I2O5 + 5CO --> 5CO2 + I2
963470
CO + 2H2  --->  CH3OH\n\nmoles H2: 2.00g/ 1g/mol = 2.00 mol\nmoles CO: 18.50g / 28g/mol = 0.6607mol\n\nyour limiting reactant is CO since for every 0.6607 moles of CO, you'll only need 1.321 moles H2. there's an excess of 0.6790 moles H2.\n\ntheoretical yield, therefore, is (1.321/2)*100 =  66.05%\n\nbased from the balanced equation, there should also be 0.6607 moles of CH3OH (MW = 32g/mol) that will be formed. so,\n\nweight of CH3OH = 0.6607mol * 32g/mol = 21.14 grams
991601
You actually form 11g CO2.\n\nPer your description, the O2 is a limiting reactant. I.e., there is a fixed amount of it and more C2H2. So, the balanced equation becomes:\n\n2C2H2 + 5O2 = 4CO2 + 2H2O  (balanced equation)\n\nx = 10 g O2 x 1 mol/32g x 4 mol CO2/5 mol O2 x 44 g CO2/mol\n\nx = 11 g CO2 formed\n\nIf you burn C2H2 to completion the equation changes to:\n\nx = 10 g C2H2 x mol/26g x 4 mol CO2/2 mol C2H2 x 44 g/mol CO2\n\nx = 33.8 g CO2 formed
303776
Gasoline + Oxygen = CO2 + H2O + NOx + SO4 + energy
694912
I do not know where you got your formula for cellulose.  Cellulose is usually considered as having variable length of anywhere from 2000 to 14000 units of -(C6H10O5)-\n\nUsing C52H84O22, combustion would be:\n\n62 O2 (g) + C52H84O22  (s) -->  52 CO2 (g) + 42 H2O (g)
41822
Simple one will be \nSunlight + Carbon dioxide (CO2) + Water (H2O)--> Glucose (C6H12O6) + Oxygen (O2).\nor \n6CO2 +12 H2O (light energy and chlorophyll) ---> C6H12O6 + 6O2\nRemember to use subscript whenever necessary.
278240
C6H12O2 + 6O2 --> 6CO2 + 6H2O\n\nis the combustion reaction for glucose.
832153
2 moles of reactant  (C2H4F2) give you 4 moles of HF\nC2H4F2 : 66 g/mol\nHF: 20g/mol\nTherefore\n79,3g  C2H4F2*(1mol  C2H4F2/66g  C2H4F2)*(4 moles HF/ 2moles C2H4F2)*(20g HF/ 1mol HF)= 48,1g HF\nThe amount of HF you are expected to get with 100% yield is 48,1g. If you got 18g, then\n(18g/48,1g)*100= 37% yield
943
the number of carbon-12 atoms in 12 grams (0.012 kg) of carbon-12 or 6.02214199 × 10^23
38679
C6H8O7\nassume a 100 gram sample. meaning there's that many grams of each.  now you divide each percent by the atomic mass of the element. this gives you the number of moles of each.\nC: 37.51/12.01 = 3.12\nH: 4.20/1.008 = 4.16\nO: 58.9/16.00 = 3.68\nnow you know the number of moles, so all you have to do is get them as whole number coefficients.  so first, divide by the lowest number this way at least one of them is a 1.\nC: 3.12/3.12 = 1\nH: 4.16/3.12 = 1.333\nO: 3.68/3.12 = 1.18\nnow you see that the H has a .333 which is a third, so you multiply them all by three to make the H a whole number too...\nC: 1.0 x 3 =   3\nH: 1.333 x 3 = 4\nO: 1.78 x 3 = 3.54\nall that's left is the O, and that's close to 3.5, so we multiply them all by two getting:\nC: 3.0 x 2 = 6\nH: 4.0 x 2 = 8\nO: 3.5 x 2 = 7\nso the emperical formula is C6H8O7 (where the numbers are subscripts)
714204
What teacher still gives exercises in such mixed units? Degrees Fahrenheit and Kelvins. Distance in feet....but the results is required in grams?\n\nThe volume of air is 16000 ft^3 or 453 m^3. The density of air is 1.293 gram/liter (I looked that up) or 1293 gram/m^3. We have therefore 5.86 x 10^5 grams of air to heat.\nThe temperature change is 30 deg F or 16.7 deg K.\nSo the total heat needed is 1.01 x 16.7 x 5.86 x 10^5 -9.88 x 10^7 Joule. \nThe combustion of methane provides 890.8 x 10^3 J/mol, which means that we need 111 mol of methane or 111 x (12+4x1) = 1775 gram of CH4\n\nThe reaction equation is CH4 + 2 O2 -> CO2 + 2 H2O.\nSo one mol of methane will produce one mol of CO2. A total of 111 mol are produced, or 111 x (12+2x16) = 4884 gram of CO2.\n\n(And all that stuff goes in the atmosphere warming up the greenhouse a little more.)
228649
CO2 + H2O + light = C6H12O6 + O2
15061
K2Cr2O7 + 2Pb(C2H3O2)2 + H2O --> 2PbCrO4 + 2HC2H3O2 + 2KC2H3O2
825261
First you must balance the equation:\n4NH3 + 5O2  -> 4NO + 6H2O\nthere are 4N, 12H and 10O on each side\nyou have 4.78 moles of O2 you need to divide it by the number of moles of O2 that react and multiply it by the number of moles of NO produced.\n4.78 / 5 x 4 = 3.824moles\nHope this helps