How do we calculate the number of grams of water formed from the reaction
2H2(g) + O2(g) 2H2O(g)
if we have 13.6 liters of O2(g) at 10 0C and 755 mm Hg mixed with 34.6 liters of H2(g) at 28 C and 724 mm Hg to produce water?
Now we can calculate the number of moles of the reactants from the ideal gas law. However our units are not what we need for we need pressure in atmospheres and temperature in Kelvin.
For oxygen, T = 10 + 273 = 283 K and P = 755 mm Hg/(760 mm Hg/atm) = 0.993 atm, so for oxygen:
noxygen = PV/RT = 0.993 atm*13.6 l/(0.0821 latm/molK)*283K) = 0.581 moles
For hydrogen, T = 28 + 273 = 301 K and P = 724 mm Hg/(760 mm Hg/atm) = 0.952 atm, so for hydrogen:
nhydrogen = PV/RT = 0.952 atm*34.6 l/((0.0821 latm/molK)*301K)= 1.33 moles.
So the ratio of hydrogen to oxygen that we actually have is 1.33/0.581 = 2.29 but the ratio that we need is given by the ratio of the stoichiometric coefficients which is 2/1 = 2 so we have too much hydrogen and thus oxygen is the limiting reagent. Thus we will use oxygen to calculate how much water the above reaction forms:
0.581 moles oxygen * (2 moles water/mole oxygen) = 1.162 moles of water formed. The gram molecular weight of water is 18.0 grams so the number of grams of water formed is
1.162 moles water * 18.0 grams/mol = 29.916 g = 30. g water
Easy! But you must practice, practice, practice.
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