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• 1. coza b  |  May 12th, 2009 at 10:58 pm

  8.
  All water evaporated = water loss = 11.88 - 7.6 = 4.28 g.
  moles of water = 4.28/18 = 0.2378 moles
  13.
  Molecular weight
  CaF2=78
  HF=20
  source: http://www.uwosh.edu/faculty_staff/kedro...
  percent yield = actual yield/theoretical yield
  theoretical yield = 2x(9/78)moles = 0.2308 (based on coefficients, 1 mole of CaF2 produce 2 moles of HF)
  actual yield = (1.76/20)moles = 0.088
  percent yield = 0.088/0.2308 = 38.12%
  16.
  Atomic weight of Fe = 56
  mol weight of Fe2O3 = 160
  4Fe + 3O2 –> 2Fe2O3
  Fe stays as Fe = 7/8×627 = 548.625
  Fe become Fe2O3 = 1/8×627 = 78.375 ==> 78.375/56 = 1.4 moles,
  Fe2O3 = 2/4 x 1.4 moles x 160 = 112 g
  total wt = 112+548.625= 660.625 g
  17.
  assume all glucose become energy+water+Co2
  (mol wt=180)
  C6H12O6 + 6O2 –> 6CO2 + 6H2O
  300 g/day
  total CO2 = 6 x (300/180) moles/day x 365 days/year x 6 billion x 44 g/mole CO2
  = 9.636 x 10^14 g/year
  23.
  Mr Fe2O3 = 160 g/mole
  1000 kg Fe = 1000/56 kmoles
  Fe2O3 needed = 1/2 x 1000/56 = 1000/112 kmoles
  Fe2O3 needed = 1000/112 x 160 = 1428.57 kg
  purity = 1428.57/2250 = 63.492%

• 2. Dani  |  May 12th, 2009 at 10:58 pm

  sorry, you should go talk to your teacher BEFORE school not before or after class or during either you lose brownie points for that you gain for before school

• 3. neonpony  |  May 12th, 2009 at 10:58 pm

  i would like to help you but i cant see the questions

• 4. ME2010  |  May 12th, 2009 at 10:58 pm

  I need some help with that stuff as well. I can do some stoichiometry, and from my understanding it’s basically converting the grams you want to moles, then multiplying that by the ratio, then multiplying it back to grams.

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