Question

If 5.03 g of Ag2O is sealed in a 75.0-mL tube filled with 760 torr of N2 gas at 32 ∘C, and the tube is heated to 330 ∘C, the Ag2O decomposes to form oxygen and silver. What is the total pressure inside the tube assuming the volume of the tube remains constant?

Answer 1

step 1: calculate initial mol of N2 gas

Given:

P = 760.0 torr

= (760.0/760) atm

= 1 atm

V = 75.0 mL

= (75.0/1000) L

= 0.075 L

T = 32.0 oC

= (32.0+273) K

= 305 K

find number of moles using:

P * V = n*R*T

1 atm * 0.075 L = n * 0.08206 atm.L/mol.K * 305 K

n = 2.997*10^-3 mol

step 2: Now calculate the mol of O2 formed

Molar mass of Ag2O,

MM = 2*MM(Ag) + 1*MM(O)

= 2*107.9 + 1*16.0

= 231.8 g/mol

mass of Ag2O = 5.03 g

mol of Ag2O = (mass)/(molar mass)

= 5.03/2.318*10^2

= 2.17*10^-2 mol

Balanced chemical equation is:

2Ag2O ---> 4 Ag + O2

According to balanced equation

mol of O2 formed = (1/2)* moles of Ag2O

= (1/2)*2.17*10^-2

= 1.085*10^-2 mol

step 3: calculate final pressure

final mol of gas = 2.997*10^-3 mol + 1.085*10^-2 mol

= 0.01385 mol

Given:

V = 75.0 mL

= (75.0/1000) L

= 0.075 L

n = 0.01385 mol

T = 330.0 oC

= (330.0+273) K

= 603 K

use:

P * V = n*R*T

P * 0.075 L = 0.0138 mol* 0.08206 atm.L/mol.K * 603 K

P = 9.1377 atm

Answer: 9.14 atm