Question

A typical room is 4.0 m long, 5.0 m wide, and 2.5 m high.

Part A) What is the total volume of nitrogen in the room assuming that the gas in the room is at STP and that air contains 21% oxygen and 79% nitrogen?

Part B) What is the total number of moles of nitrogen?

Answer #1

Volume of room = length×height×width

We know that 1m = 100cm

So

Length = 4.0m = 400cm

Height= 2.5m = 250cm

Width = 5m = 500cm

Volume of room = 400cm×500cm×250cm = 50000000cm^{3}

Part(a)

volume of nitrogen gass = 79% of total volume of room

Volume of nitrogen gas = (50000000cm^{3})×79/100 =
39500000cm^{3}

because

1L = 1000cm^{3}

So, 1cm^{3} = (1/1000)L

Hence, 39500000cm^{3} = (39500000/1000)L = 39500L

Volume of nitrogen = 39500Liter

Part (b)

We know that at STP, volume of one mole gas is 22.5L

22.5L = 1mol

1L= (1/22.5)mol

39500L= (39500/22.5)L = 1755.555mol

Number of moles of nitrogen gass = 1755.555mol

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