Calculate the volumes of air and hydrogen gas

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SUMMARY

This discussion focuses on calculating the volumes of air and hydrogen gas required to produce 1,000 kg of pure molybdenum from molybdenite (MoS2). The process involves oxidizing MoS2 to molybdenum trioxide (MoO3) and then reducing MoO3 to metallic molybdenum using hydrogen gas. Key calculations include converting mass to moles, determining the moles of hydrogen needed based on stoichiometry, and adjusting for the fact that air contains only 21% oxygen. The ideal gas law (PV=nRT) is utilized to convert moles into volumes at specified conditions.

PREREQUISITES
  • Understanding of stoichiometry in chemical reactions
  • Familiarity with the ideal gas law (PV=nRT)
  • Knowledge of molybdenum chemistry, specifically the reduction of MoO3
  • Basic concepts of gas composition, particularly the percentage of oxygen in air
NEXT STEPS
  • Calculate molar mass and moles of molybdenum from 1,000 kg
  • Learn to apply the ideal gas law for gas volume calculations
  • Study the reduction reactions of metal oxides using hydrogen
  • Explore the implications of gas composition on reaction yields
USEFUL FOR

Chemistry students, chemical engineers, and professionals involved in metallurgical processes or gas reaction calculations will benefit from this discussion.

Lilly
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If anyone knows how to do this problem, could you please give me some
hints?? I don’t need the answer just some help as to where to start...

Metallic molybdenum can be produced from the mineral molybdenite, MoS2.
The mineral is first oxidized in air to molybdenum trioxide and sulfur dioxide.
Molybdenum trioxide is then reduced to metallic molybdenum using hydrogen
gas. The balanced equations are:

MoS2 (g) + 7/2 O2 (g) -------> MoO3 (s) + 2SO2 (g)

MoO3 (s) + 3H2 (g) ------------> Mo (s) + 3H2O (l)

Calculate the volumes of air and hydrogen gas at 17 deg. C and 1.00 atm
that are necessary to produce 1.00 x 10^3 kg of pure molybdenum from
MoS2 . Assume air contains 21% oxygen by volume and assume 100% yield
for each reaction.
 
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Hello,

1) One ton of pure molybdenum is not a useful definition, it's better to convert it into moles.

2) As one mole of molybdenum is produced by three moles of hydrogen, you can find how many moles of H2 is needed, and then convert it into volume by P*V=n*R*T.

3) The mole of molybdenum is the same as molybdenum(VI)oxide; just put this number in the first reaction, and multiply it by 3,5 to obtain how many moles of pure oxygen is needed. Then convert it into volumes.

4) If 100% of oxygen is present in the environment, the volume you find will be enough, but only 21% of oxygen is present in the air. You'll have to divide something to learn the actual volume.

Hope these help.

chem_tr
 
Thank you for your help chem_tr, for this problem as well as the one on effusion rates! Helped a lot. :)
 

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