Question

A steel cylinder with a volume of 15.0 L is filled with 69.4 g of nitrogen gas at 25∘C.

Part A

What is the pressure, in atmospheres, of the N2 gas in the cylinder?

Express your answer to three significant figures with the appropriate units.

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Part B

How many liters of H2 gas can be produced at 0 ∘C and 1.00 atm (STP) from 44.5 g of Zn?
Zn(s)+2HCl(aq)→ZnCl2(aq)+H2(g)

Express your answer with the appropriate units.

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A gas mixture containing oxygen, nitrogen, and neon exerts a total pressure of 1.40 atm .

Part C

If helium added to the mixture increases the pressure to 1.70 atm , what is the partial pressure (atm) of the helium?

Answer 1

A)

Molar mass of N2 = 28.02 g/mol

mass(N2)= 69.4 g

use:

number of mol of N2,

n = mass of N2/molar mass of N2

=(69.4 g)/(28.02 g/mol)

= 2.477 mol

Given:

V = 15.0 L

n = 2.4768 mol

T = 25.0 oC

= (25.0+273) K

= 298 K

use:

P * V = n*R*T

P * 15 L = 2.4768 mol* 0.08206 atm.L/mol.K * 298 K

P = 4.04 atm

Answer: 4.04 atm

B)

Molar mass of Zn = 65.38 g/mol

mass of Zn = 44.5 g

mol of Zn = (mass)/(molar mass)

= 44.5/65.38

= 0.6806 mol

According to balanced equation

mol of H2 formed = moles of Zn

= 0.6806 mol

Given:

P = 1.0 atm

n = 0.6806 mol

T = 0.0 oC

= (0.0+273) K

= 273 K

use:

P * V = n*R*T

1 atm * V = 0.6806 mol* 0.08206 atm.L/mol.K * 273 K

V = 15.2471 L

Answer: 15.2 L

C)

p(He) = p(after He added) - p(before He added)

= 1.70 atm - 1.40 atm

= 0.30 atm

Answer: 0.30 atm