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

Chemistry

1 answer:

sineoko [7]2 years ago

7 0

Answer: 17 grams

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number $6.023\times 10^{23}$ 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= ?

$n=\frac{PV}{RT}=\frac{1\times 5.9}{0.0821\times 273}=0.26moles$

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

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

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