Hi, I just need a little help with this homework problem. If you could show the steps that would be awesome. Thanks a lot, all the help is greatly appreciated.
How many liters of water vapor can be made from 55.0 grams of oxygen gas and an excess of hydrogen gas at a pressure of 12.4 atm and a temperature of 85.4 degrees Celsius?
How many liters of water vapor can be made from 55.0 grams of oxygen gas and an excess of hydrogen gas at a pressure of 12.4 atm and a temperature of 85.4 degrees Celsius?
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H2 + O2 ---> 2H20
Since there is an excess of H2, O2 is the limiting reactant and will determine the number of moles of water vapor in the product.
moles, O2 = mass / molecular weight = 55.0 g / (32.0 g/mol) = 1.71875 moles of O2
1.71875 moles O2 x (2 moles H2O / 1 mole O2) = 3.4375 moles of H2O
We can now solve for the volume (liters of water vapor) by using the ideal gas equation:
pV = nRT
(12.4 atm)(V) = (3.4375 moles)(.082057 L atm / K mol)(358.4 K)
V = 8.15 L
Hope this helps...good luck!
Since there is an excess of H2, O2 is the limiting reactant and will determine the number of moles of water vapor in the product.
moles, O2 = mass / molecular weight = 55.0 g / (32.0 g/mol) = 1.71875 moles of O2
1.71875 moles O2 x (2 moles H2O / 1 mole O2) = 3.4375 moles of H2O
We can now solve for the volume (liters of water vapor) by using the ideal gas equation:
pV = nRT
(12.4 atm)(V) = (3.4375 moles)(.082057 L atm / K mol)(358.4 K)
V = 8.15 L
Hope this helps...good luck!