asked 106k views
1 vote
93.0 mL of O2 gas is collected over water at 0.930 atm and 10.0 C.what would be the volume of this dry gas at STP

1 Answer

2 votes
Let us assume that the oxygen behaves as an ideal gas such that we can use the ideal gas equation to solve for the number of moles of O2.
PV = nRT ; n = PV/RT
Substituting the known values,
n = (0.930 atm)(93/1000 L) / (0.0821 L.atm/mol.K)(10 + 273.15K)
n = 3.72 x 10^-3 mols
At STP, the volume of each mol of gas is equal to 22.4 L.
volume = (3.73 x 10^-3 mols) x (22.4 L/1 mol)
volume = 0.0833 L or 83.34 mL
answered
User Brezhnews
by
9.6k points
Welcome to Qamnty — a place to ask, share, and grow together. Join our community and get real answers from real people.