Liquid octane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 6.85 g of octane is mixed with 13. g of oxygen. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. Round your answer to significant digits.

  remember that hf is only the energy of one photon. not two. not three. one.  1) 5.12 * 10^3 photons  find the energy of one photon, then determine how many photons are needed to reach the sensitivity.  e = hf  = (6.626 * 10^-34 js)(5.95 * 10^19 hz)  = 3.94247 * 10^-14 j per photon  (2.02 * 10^-10 j) / (3.94247 * 10^-14 j/photon) = 5123.69149 photons  round to three sig figs.  2) 5.77 * 10^-17 j  energy of pulse = energy per photon * number of photons  = hfn  = (6.626 * 10^-34 js)(1.66 * 10^13 hz)(5250)  = 5.774559 * 10^-17 j  round to three sig figs.  3) 4.07 * 10^8 kj/mol  10^12 pm = 1 m  c = fλ  f = c/λ  e = hf  = hc/λ  = [(6.626 * 10^-34 js)(3.00 * 10^8 m/s) / (2.94 * 10^-12 m)](1 kj/1000 j)(6.02 * 10^23 photons/mol)  = 40702571.4 kj/mol  round to three sig figs.  4) 9.94 * 10^-1 m  e = hc/λ  λ = hc / e  = (6.626 * 10^-34 js)(3.00 * 10^8 m/s) / (2.00 * 10^-25 j)  = 0.9939 m  round to three sig figs.  5) 9.06 * 10^3 photons  energy of pulse = energy per photon * number of photons  number of photons = energy of pulse / energy per photon  = energy of pulse / (hf)  = (3.87 * 10^-13 j) / [(6.626 * 10^-34 js)(6.45 * 10^16 hz)]  = 9055.23695 photons  round to three sig figs.  6) 2.78 * 10^-15 j  this is problem 1, except in reverse.  sensitivity = minumum number of photons * energy per photon  = (5930)(6.626 * 10^-34 js)(7.077 * 10^14 hz)  = 2.78070758 × 10^-15 j  round to three sig figs.

The answer is ax3. hope this

Unbalanced equation: fe2o3+ cofe + co21 fe2o3 + 3 co = 2 fe(s) + 3 co212.8g fe2o3 x (1 mol fe2o3 / 159.7g fe2o3) = 0.08015 mol fe2o311.5g co x (1 mol co / 28.01g co) = 0.4106 mol cofrom the balanced equation, 1 fe2o3 reacts with 3 cosince we have 0.08015 mol fe2o3 we therefore need0.08015 mol fe2o3 x (3 mol co / 1 mol fe2o3) = 0.2404 mol cosince we have 0.4106 mol co but only need 0.2404 to completely consume all thefe2o3, we have xs (excess) co and fe2o3 is therefore the limiting reagentfrom the balanced equation, 1 fe2o3 = 2 fe(s)0.08015 mol fe x (2 mol fe(s) / 1 mol fe2o3) = 0.1603 mol fe(s)theoretical mass of fe .1603 mol fe x (55.84g fe / mole fe) = 8.951g fe% yield = (6.10g / 8.951g) x 100% = 68.1%i didn't do this work, so i am not one hundred percent sure if it is correct  seems like a reliable website, 

In science and engineering, the parts-per notation is a set of pseudo-units to describe small values of miscellaneous dimensionless quantities, e.g. mole fraction or mass fraction. ~hope dis