assuming stp and a 21% oxygen 79% ratio in air
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The volume of the room is 3.5m X 6 m X 2 m = 42 m^3
The volume of the room in L = 42 m^3 X 1000 L/m^3 = 4.2X10^4 L
Now, at STP, 1 mole of any gas occupies a volume of 22.4 L. So, the moles of gas in the room is:
4.2X10^4 L X 1 mol/22.4L = 1875 moles. Since O2 makes up 21% of air,
1875 mol X 0.21 = 394 moles O2
Mass O2 = 394 moles X 32 g/mol = 1.26X10^4 grams O2 in the room
The volume of the room in L = 42 m^3 X 1000 L/m^3 = 4.2X10^4 L
Now, at STP, 1 mole of any gas occupies a volume of 22.4 L. So, the moles of gas in the room is:
4.2X10^4 L X 1 mol/22.4L = 1875 moles. Since O2 makes up 21% of air,
1875 mol X 0.21 = 394 moles O2
Mass O2 = 394 moles X 32 g/mol = 1.26X10^4 grams O2 in the room
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8.82?
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what are the units of measurement of the room 3.5 ? long