Question

A sample containing 4.20 g of O2 gas has an initial volume of 20.5 L . What is the final volume, in liters, when each of the following occurs and pressure and temperature do not change?A sample of 2.25 g of O2 is removed.

Answer 1

V2 = 10.95 L

Step-by-step explanation

Given:

mass 1 = 4.20 g

mass 2 = 2.25 g

volume 1 = 20.5 L

Required: volume 2

Solution

Step 1: Find the number of moles of mass 1 and 2.

For mass 1:

n = m/M where n is the moles, m is the mass and M is the molar mass

n = 4.20g/31.998g/mol

n = 0.131 mol

For mass 2:

n = 2.25g/31.998g/mol

n = 0.070 mol

Step 2: For this problem, we will use the ideal gas law equation

PV = nRT

However, P, V and R are constant

Therefore

`(n_1)/(V_1)\text{ = }(n_2)/(V_2)`

V2 = (V1n2)/n1

V2 = (20.5L x 0.070 mol)/0.131 mol

V2 = 10.95 L