Asked by Eileen
Oxygen gas reacts with powdered aluminum according to the following reaction:
4Al(s)+3 O2(g) --> 2 Al2O3 (s)
How many liters of O2 gas, measured at 789 mm Hg and 35 C, are required to completely react with 2.2 mol of Al?
4Al(s)+3 O2(g) --> 2 Al2O3 (s)
How many liters of O2 gas, measured at 789 mm Hg and 35 C, are required to completely react with 2.2 mol of Al?
Answers
Answered by
Damon
for every 4 mols of Al you need 3 mols of O2
so
mols O2 = (3/4)mols Al = .75 *2.2 = 1.65 mols
Now do P V = n R T with n = 1.65 mol
so
mols O2 = (3/4)mols Al = .75 *2.2 = 1.65 mols
Now do P V = n R T with n = 1.65 mol
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