Question

Calculate the grams of so2 gas present at stp in a 5.9 l container. (r = 0.0821 l·atm/k·mol)

Answer 1

Answer: 17 grams

Step-by-step explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number
`6.023* 10^(23)` of particles.

Standard condition of temperature (STP) is 273 K and atmospheric pressure is 1 atm respectively.

According to the ideal gas equation:

`PV=nRT`

P = Pressure of the gas = 1 atm

V= Volume of the gas = 5.9 L

T= Temperature of the gas = 273 K

R= Gas constant = 0.0821 atmL/K mol

n= moles of gas= ?

`n=(PV)/(RT)=(1* 5.9)/(0.0821* 273)=0.26moles`

Mass of
`SO_2=moles* {\text {Molar mass}}=0.26* 64.066 g/mol=17g`

Thus 17 g
`SO_2` gas is present at STP in a 5.9 L container.