what is the number of moles for CO^2?
365485
I assume you mean molecular weight.\n\nSince carbon weights 12 grams, and oxygen weighs 16:\n12+(16*2) = 44 grams.\n\nYou have to give the amount of CO^2, in ML to get Moles.\n\nSo generically speaking, 1 mol of CO^2 weighs 44 grams, 2 mol weights 88 grams, and so on.\n\nTo calculate the amount of atoms in it, you would have to times it with Avogadro's number. (6.0221415 × 10)
1012524
2C8H18 +   25O2  --->   16CO2 + 18H2O\n\n2 molecules of octane (which is not the only component of gasoline, and octane rating has nothing to do with how much octane is actually present) react with 25 molecules of oxygen to produce 16 molecules of carbon dioxide and 18 molecules of water.
670103
Assuming that NO2 is an ideal gas is these conditions (which is a reasonable assumption at such a low P), the number of moles into that flask is: n = PV/(RT) where P = 432 mmHg = 0.568 atm, V = 0.5 L, T = 60 C = 333 K, R = 0.082 L atm/(mol K).\nThen: n = 0.01 mol and, taking into account the Avogadro Number (in number of particles/mol) the number of molecules is NM = 0.01*6.022x10^23 = 6.022x10^21.
896682
Place a 2 in front of the HCl to balance the equation. Then you can see that for one mole of calcium carbonate, you get 1 mole of carbon dioxide.
1167519
there would be approximately 2.648*10^24 acetone molecules.  This was done by first finding the number of gram in the volume by multiplying volume by the density.  This would give the number of grams which is then converted to moles.  Then you can use avagadro's number to multiply the number of moles to find the number of molecules.  The long equation is below\n325 ml * 0.7857 g/ml * 1/58.0798 mol/g * 6.0221 X 10^23 moleucles/mol\n\n0.7857 is the density of acetone\nacetone has a molecular weight of 58.0798 g/mol \nand avagadros number is approx. 6.0221 * 10^23\n*Note you must divide by the molecular weight to get the correct units at the end
961444
We will need to find if there is a limiting material.\n\nCO + 2H2 ------> CH3OH\n\n(1.61 g H2)*(1 mole H2/2.016 g) = 0.7986 mole H2\n\n(0.7986 mole H2)*(0.5 mole CH3OH/mole H2) = 0.3993 mole CH3OH at most\n\n(15.95 g CO)*(1 mole CO/28g) = 0.5696 mole CO\n\n(0.5696 mole CO)*(1 mole CH3OH/1mole CO) = 0.5696 mole CH3OH at most\n\nH2 is the limiting reagent, therefore:\n0.3993 mole CH3OH * (32.032 g/mole CH3OH) = 12.79 g\n\nPlease check for errors, I went by memory on what the atomic and molecular weights were, you may want to check this.
839940
1. 2C6H6 + 15O2 --> 12CO2 + 6HOH\n\n2. C12H22O11 + 12O2 --> 12CO2 + 11HOH\n\n3. CH3COCH3 + 4O2 --> 3CO2 + 3HOH\n\n4. C2H5OC2H5 + 6O2 --> 4CO2 + 5HOH\n\n5. C4H9OH + 6O2 --> 4CO2 + 5HOH
764973
(0.09/6.022)*(22.414/6.022) atm\n\n0.09 /6.022= n= no. of moles (n=no. of molecules/avogadro no.)\navogadro no. =6.022*10^23
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).
804072
220g\n\n32g of 02 is one mole\n160/32=5\nand you have a 1mole to 1mole ration in the reaction so,\n12+16+16=44g in one mole of CO2\n44 x 5=220g
363082
2Ag+ + CO_33-2 -> Ag_2CO_3
841680
1) determine the mass of carbon, hydrogen and oxygen.\ncarbon: 5.752 g x (12 / 44) = 1.569 g\nhydrogen: 2.354 g x (2 / 18) = 0.262 g\noxygen: 2.356 - 1.569 - .262 = 0.525 g\n\n2) convert to moles.\ncarbon: 1.569  / 12  = 0.131 mol\nhydrogen: 0.262  / 1  = 0.262 mol\noxygen: 0.525 / 16 = 0.0328 mol\n\n3) divide each molar amount by the lowest value\ncarbon: 0.131  / 0.0328  = 4\nhydrogen: 0.262 l / 0.0328  = 8\noxygen: 0.0328 / 0.0328 = 1\n\nWe have now arrived at the answer: the empirical formula of the substance is C4H8O
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.
739541
CH4 + 2O2-> CO2 + 2H2O
641455
Least C6H6 with like .5 moles\nThen 1 mole of CH4 and the 44 g CO2 both have same\nAnd then the 2 mole Br2
591191
if i remember my chemistry right, a mole of gas at stp is 22.4 liters, so the answer should be E), 2 moles
800696
19) The number of moles of CO in 86g is: 86 / 28 = 3.07 moles.\n\nCO + 2H2 ----> CH3OH\n\nFrom this equation, 1 mole CO, when reacting with H2, is supposed to produce 1 mole CH3OH.\n\nTherefore, 3.07 moles of CO should produce 3.07 moles of CH3OH, which equals 3.07 * (12 + 4 + 16) = 98.29g.\n\nIn fact, we've got only 68g, hence, the percent yield of CH3OH is:\n(68 / 98.29) * 100% = 69.19%\n\nThe next question is essentially the same.