Question

In a container with volume of 25.0 L, there are 40 g of CH4 gas. If the number of gas is reduced to 15.0 L, what is the new amount inmole?

Answer 1

Answer

1.50 mol

Explanation

Given:

Initial volume, V₁ = 25.0 L

Mass of CH4 gas in 25.0 L container = 40 g

Final volume, V₂ = 15.0 L

From the Periodic Table; molar mass of CH4 = 16.04 g/mol

What to find:

The new amount in mole.

Step-by-step solution:

According to Avogadro’s law: For a confined gas, the volume (V) and number of moles (n) are directly proportional if the pressure and temperature both remain constant. That is:

$\frac{V_1}{n_1}=\frac{V_2}{n_2}$

n₁ = Mass/Molar mass = (40.0g/16.04 g/mol) = 2.493765586 mol

n₂ is the new amount in mole and can be calculated as follows:

$\begin{gathered} \frac{25.0\text{ L}}{2.493765586\text{ mol}}=\frac{15.0\text{ L}}{n_2} \\ \text{Cross multiply} \\ n_2\times25.0\text{ L }=15.0\text{ L }\times2.493765586\text{ mol} \\ \text{Divide both sides by 25.0 L} \\ \frac{n_2\times25.0\text{ L}}{25.0\text{ L}}=\frac{15.0\text{ L }\times2.493765586\text{ mol}}{25.0\text{ L}} \\ n_2=1.496259352\text{ mol} \\ To\text{ 3 significant digits} \\ n_2=1.50\text{ mol} \end{gathered}$

The new amount in moles is 1.50 moles