what mass of carbon monoxide must react w/oxygen to produce .69 of carbon dioxide?
116688
from the equation: 2CO + O2 -----> 2CO2,\n\n0.69 g CO2 x (1 mol CO2/ 44 g CO2) x (2 mol CO/ 2 mol CO2) x (28 g CO/ 1 mol CO)\n\n\nthe answer is 0.4391 g of CO needs to react with O2 to produce 0.69 g of CO2
764912
If I understand correctly what you need then this is what you do:\n(by the way we will take He as 83.7 because it is over 100% other wise.)\nmultiply the molecular or atomic weight of each species by the fraction (percent divided by 100).\n\nMW(CO2) = 44\nMW(N2) = 28\nAW(He) = 4\n\n44*0.11 + 28*0.053 + 4*0.837 = 9.67 g/mol\n\nThis is the equivalent "molar mass" of the mixture.
446321
Hmmm, a multi-step problem!  Love those!\n\nFirst, you will need to know how many moles of oxygen are generated.  For that, you will need to use the Ideal Gas Law:\n\nPV = nRT\n\nYou need to solve for n = PV/RT, and since you know the volume, pressure, and temperature, and the value of R is a constant, this is an easy calculation.\n\nThen, from the equation:  \n\nKClO3 ===> KCl + 1.5 O2\n\nyou can work backwards to determine the number of moles of KClO3 you started with (one mole of KClO3 generates one and a half moles of oxygen, thus the number of moles of KClO3 you started with will be 2/3 of the number of moles of oxygen generated).  Knowing how many moles of KClO3 you had, you can then determine the starting mass of the KClO3, then find out what percentage that was of your original starting mixture.\n\nGood Luck!
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
977270
first you need a balanced equation.\n\n2Mg+O2===>2MgO\n\nNext convert everything given to mols.\n\n10.000 g Mg X 1 mol Mg / 24.30 g Mg (NOTE: this number is from the periodic table of the Elements) = 0.4115 mol Mg\n\n5.267 g O2 X 1 mol O2/ 32.00 g O2 (this is the molar mass of O2)= 0.1646 mol O2\n\nAt this point pick one of the mols you calculated.\n\n0.1646 mol O2 X 2 mol Mg (from balanced equation) / 1 mol O2 (from balanced equation)(for every 2 mol of Mg there is 1 mol of O2)= 0.3292 mol Mg needed to react with all of O2\n\nSince more Mg is available then needed (0.4115 mol > 0.3292 mol) this implies that O2 is the limiting factor. \n\nUse the limiting factor (O2) to get the mass of the MgO\n\n0.1646 mol O2 X 2 mol MgO / 1 mol O2= 0.3292 mol MgO\n\n0.3292 mol MgO X 40.30 g MgO/ 1 mol MgO = 13.27 g MgO\n\nessentially for all the other ones you are doing the same thing as before but with different numbers.
254672
You answer this question by formula weight ratios.  One mole of Na2CO3.10H2O has a mass of 286.14 g.  It also contains 13 moles of oxygen (3 in the carbonate + 10x1 each water), which is 13x15.9994 = 207.99 g.\n\nThe mass percent oxygen is then (amt in question / amt total)x100% = (207.99/286.14)x100% = 72.688%, or 72.7% to the nearest tenth.
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
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
363430
Carbon with 3 bonds to oxygen\n\nC-O\n\nwith 3 of the - signs\n\nIt makes carbon monoxide
541101
First you need the equation for the reaction of CO2 with phenyl magnesium bromide to get the mole ratio. (i get one to one.)\n\nSecond, calculate the number of millimoles of CO2 required to react.\n\nLastly, multiply the number of millimoles of CO2 by its molar mass. I get 88 mg.
844628
a) 53.4\n34gCO  x 1mol CO/28gCO x 2molCO2/2molCO x44gCO2/1molCO2 = 53.4gCO2\nThe 2 moles CO cancel the 2 moles CO2 (from the balanced equation).
843441
The answer is c.\nC12H22O11+12O2-->12CO2+11H2O
410173
This formula might help\n\nchange in S=(m1)(c1)(ln(Tf/Tc))+(m2)(c2)(ln(Tf/Tc))\nm1 & m2=substance mass in this case oxygyn gass (O2)=15.999(2)\nc1 & c2 are spcific heats\nTc =cold temp\nTh = Hot Temp\nTf=((m1)(c1)(Tc)+(m2)(c2)(Th))/((m1)(c1)+(m2)(c2))\n\nI'm assuming m^3/kg is the pressure (Volume/Mass) ???\n\nhope this helps!!
445028
Carbondioxide gas, Oxygen gas and Nitrogen gas
929547
I find 16.4% \nHere we go:\n1) As air is a mixture of O2 and N2 two reactions are taking place:\n2 Mg + O2 -> 2 MgO\n3 Mg + N2 -> Mg3N2 \nThere are 4 parts of N2 for one part of O2, so the second reaction takes place four times as frequently as the first. Effectively we have\n14 Mg + O2 + 4 N2 -> 2 MgO + 4 Mg3N2.\n\n2) The molar mass of MgO is 40.3 gram. In MgO the magnesium makes up 24.3/(24.3+16.0) = 60.3 % of the mass.\nThe molar mass of Mg3N2 is 100.9 gram, the magnesium makes up 3*24.3/(3*24.3 + 2*14.0) =  72.2 % of the mass. \n\n3) Assume you have 1000 gram of reaction product. If it is pure MgO you will find 603 gram of Mg. But in the mixture you will have  \ntwo moles of Mg2N3 (201.8 g) for every mole of MgO (40.3 gram).\nIn 1000 gram of mixture, 833.5 gram of  Mg2N3 and 166.5 gram of MgO. This gives 72.2% * 833.5 + 60.3% * 166.5 = 702 gram of Mg.\n\n4) Finally: the error is (702-603)/603 = 16.4%
968504
The balanced reaction is:\n3C + 2SO2  ---->  CS2  +  2CO2\n\nNotice that in the balanced reaction that the number of moles of sulfur dioxide that react and the moles of carbon dioxide produced are the same.\n\nSince the molar mass of SO2 is 64g/mol and you are starting with 8.00 grams, this means that you are starting with 1/8 of a mole of sulfur dioxide.  Since carbon dioxide is produced at the same amount, 1/8 of a mole of CO2 will be produced.
851681
Write the balanced equation - you need the mole ratios.\n\nChange 1g hexane to moles of hexane.\n\nSee how many moles of oxygen are USED and how many moles of  CO2 and H2O are formed.\n\nYour total number of moles in the reaction mixture would then be:\n\nmoles of Oxygen NOT used + moles of CO2 (we will assume the water formed consenses to a liquid at 20 degrees).\n\nUse the Ideal Gas Law to figure out the volume.\n\nPV = nRT\n\nwhere P = given pressure - water vapor pressure\nwhere V = your unknown\nwhere n = your total moles of gas\nwhere R = the appropriate Universal Gas Law constant\nwhere T = the temperature in KELVIN