If a car produces 88 grams of CO2, how many grams of O2 are used up in the reaction?
815144
64 grams... multiply 88 grams by the ratio of molar masses of O2 to CO2 to get  the result
876161
You're right - as written it's impossible to balance.  But if you write\n\na(Ag2CO3) --> b(Ag2O) + c(CO2) + d(O2) you can balance it with the values \na=1, b=1, c=1, d=0.    Meaning that the reaction doesn't generate any oxygen.  The carbonate falls apart into the oxide and gaseous CO2.
865243
C3H6 +  9/2 O2 ---> 3 CO2 + 3 H2O\nassumed oxygen is ideal gas, so\nPV=nRT\nR = 82.057 cm3.atm/gmol.K\nV = 15 l = 15000 cm3 ; T = 37 C = 310 K\nn (mol) O2 = PV/RT = 2.4*15000/82.057*310 = 1.415 gmol\nfrom balanced reaction CO2 = 3/(9/2) * 1.415 gmol = 0.943 gmol\n                                         = 0.943 * 44 = 41.507 gram
866626
O3 = 16g x 3 = 48g/mol.\n 1 Mole O3 = 48g  (÷ 4)\n 0∙02083333...Mole = 1g    (x 0∙740)\n 0∙014516666.Mole = 0∙740g\n\nNO = 14g + 16g = 30g/Mol.\n 1 Mole NO = 30g\n 0∙03333...Mole = 1g\n 0∙022333.. Mole = 0∙670g\n\n If the O3 splits to give O2 + O,\nthen the reaction is 1:1, and O3 is the limiting reactant.\n\nExcess NO: \n = 0∙022333.. Mole of NO  -  0∙014516666.Mole of O3\n = 0∙0068666...Moles of NO.\n\n If the O3 splits to give O2 + O, and splits again to give O + O + O, then the reaction is 3:1, and NO is the limiting reactant.\n\nExcess O3:\n = 2/3 x  0∙014516666.Mole \n = 0∙009677777....Moles of O3.
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.
303776
Gasoline + Oxygen = CO2 + H2O + NOx + SO4 + energy
873085
88.02 g of CO2 are equivalent to 88.02/44 moles of CO2 = 2 moles\nAccording to the equation 1 mol LiOH reacts with 1 mol CO2,\n\n2 moles CO2 will require 2 mol LiOH.\n\nTo find out the gram amount multiply 2 moles of LiOH by its molecular weight\n\n2 x 23.95 = 47.9 grams LiOH.
887756
Basic balancing of chemical reactions. You can always add CO2 O2 and H2O to either side. Start with the Carbon. balance the Oxygen last since it is contained in CO2 and H2O.\n\n1) 2 C6H6 + 15 O2 ==> 12 CO2 + 6 H2O\n2) C12H22011 + 12 O2 ==> 12 CO2 + 11 H20\n3) C3H6O + 4 O2 ==> 3 CO2 + 3 H2O\n4) C4H10O + 6 O2 ==> 4 CO2 + 5 H2O\n5) C4H10O + 6 O2 ==> 4CO2 + 5 H2O
930613
CHEMICAL EQUATION\n\nC3H8 + 5O2 --> 3CO2 + 4H2O\n\n117g C3H8 * (1mol C3H8 / 44.09652g C3H8) = 2.65mol C3H8\n2.65mol C3H8 * (5 mol O2 / 1 mol C3H8) = 13.3 mol O2\n\n14.0°C + 273.15 = 287.15K\n\n 100.0 kPa (1 atm / 101.325 kPa) = .9896 atm\n\nPV = nRT\nV=nRT/P\n\nV= (13.3 mol*.0821L atm mol-1 K-1 * 287.15K) / (.9896 atm) = 317 L O2
702103
1) Determine how many moles of C3H8 we have:\n147 g/ (3*12.01+8*1.008) =3.33 mol\n\n2) Write up to reaction to determine the amount of O2 in moles:\nC3H8 + 5 O2 -> 3 CO2 + 4 H2O \nWe need 5*3.33 = 16.67 mol of O2\n\n3) The volume of 1 mol gas is 22.4 liter at STP. Determine at the given conditions:\nV(1mol) = 22.4 * (303/273) = 24.86 liter \n\nTherefore 16.67 mol = 414 liter = 0.414 m^3
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
519609
They use methanol.  Probably about 100 gallons or so per car.
381846
Nearly everything you buy or use cases uses energy to make it. To make energy (elecricity, gas, making your car go using petrol) creates carbon dioxide (CO2).\n\nTrees, plants and algae use up CO2 to make oxygen. So to reduce the amount of CO2 in the atmosphere (which is what I think you are asking) we all need to have enough plants change CO2 into Oxygen quicker than we're producing CO2.\n\nSo the best way to to practice "Waste Minimisation". The "waste heirarchy" is used to decide how to achieve this and can be applied to everything you use. It is...\n\nAvoidance \nRe-use \nRecycling \nRecovery of energy \nTreatment \nContainment \nDisposal. \n\nWhat this means is that the first thing you should try to do is avoid using it where possible. SO walk instead of using the car for example.\n\nIf you must use the product then try and buy something that you can re-use and re-use it as often os possible and so-on so that the LAST thing you do is dispose of something. By doing this you will make a simple and personal step to avoid waste and also avoid wasting energy and producing CO2.\n\nThe other thing you can do is to plant trees! They're great at using up carbon dioxide.
957905
the balanced reaction\nC12H22O11 + 12 O2 ---> 12 CO2 + 11 H2O\nso the right answer are:\nc. \nand\ne.
698109
1 mol the gas occupies 24dm3\n\n1 mol of the gas is 44 grams\n   (C=12, O= 16, 12+16+16=44)\n\n1 kg = 1000 g\n\n1000 g * 1mol/44g = 22.727 mol of CO2\n\n22.727 mol CO2 * 24dm3/1mol = 545.448 dm3 of CO2
50834
first, u should find the no. of moles in 11 grams of CO2 gas\n\n1 mole = 44g\n\ntherefore, 11g = 11 divided by 44\nwhich is equal to 0.25 moles\n\n1 mole of co2 gas occupies 24000cm cubed.\n0.25 moles of co2 occupies x cm cubed\n\nx = o.25 times 24000\n\nwhich is equal to 6000 cm3\n\nthats the volume...\n\nby da way, r u kashyap from canada? my email id harshini317. contact me if u r kashyap sripada from canada
712891
In one experiment 32.40 g of NO is mixed with 19.20 g of O2.\n\n\nThe reaction goes: 2NO + O2 -> 2NO2\n\n\nFirst, you have to find out which is the limiting reactant.\n\nno of moles of NO\n= mass / molar mass\n= 32.40 / (14 + 16)\n= 32.40 / 30\n= 1.08\n\nno of moles of O2\n= mass / molar mass\n= 19.20 / (2 x 16)\n= 0.6\n\nGoing by the equation, there must be twice the no of moles of NO than O2. However, the no of moles of O2 supplied is more than half of the no of moles of NO. Thus, O2 is in excess and NO is the limiting reactant.\n\nWorking with the no of moles of NO,\nno of moles of NO2\n= no of moles of NO\n= 1.08\n\nmass of NO2\n= (no of moles) x (molar mass of NO2)\n= 1.08 x (14 + 2 x 16)\n= 1.08 x 46\n= 49.68g\n\nAnd there you have it! :D