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2 years ago

What is the pressure exerted by 0.122 mol oxygen gas in a 1.50-L in a container at room temperature

Chemistry

1 answer:

dem82 [27]2 years ago

7 0

Answer:

The answer to your question is P = 1.95 atm

Explanation:

Data

Pressure = P = ?

number of moles = n = 0.122

Volume = V = 1.5 l

Temperature = T = 20°C

Constant of ideal gases = 0.082 atm l/mol°K

Process

1.- Convert temperature to °K

T = 20 + 273

T = 293°K

2.- Write the formula of ideal gases

PV = nRT

-Solve for P

P = nRT/V

3.- Substitution

P = (0.122 x 0.082 x 293) / 1.5

4.- Simplification

P = 2.93 / 1.5

5.- Result

P = 1.95 atm

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<h3>Further explanation</h3>

Given

Reaction

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Required

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Solution

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b. 15 L NH3 at STP ( 1mol = 22.4 L)

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mol ratio NH3 : H2O from equation : 4 : 6, so mol H2O :

$\tt \dfrac{6}{4}\times 0.67=1$

mass H2O(MW = 18 g/mol) :

$\tt mass=mol\times MW=1\times 18=18~g$

c. mol NO at 1273 K and 1 atm :

$\tt n=\dfrac{PV}{RT}=\dfrac{1\times 35.5}{0.08206\times 1273}=0.34$

mol ratio of NO : O2 = 4 : 5, so mol O2 :

$\tt \dfrac{5}{4}\times 0.34=0.425$

Volume O2 at STP :

$\tt 0.425\times 22.4=9.52~L$

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