Question

A container of acetylene gas (C2H2) holds 3.86mol of acetylene and has a volume of 39.2L. If there were a leak and the volume were to decrease to 17.3L at the same pressure and temperature, how many moles of acetylene would have been lost from the container?

Answer 1

Answer:

2.16 moles are lost from the container

Explanation:

Step 1: Data given

Moles of acetylene = 3.86 moles

Volume = 39.2 L

The volume reduces to 17.3 L

Temperature and pressure is constant

Step 2: Calculate the new number of moles acetylene

V1/ n1 = V2/ n2

⇒with V1 = the initial volume in the container = 39.2 L

⇒with n1 = the initial number of moles acetylene = 3.86 moles

⇒with V2 = the new volume in the container after the leak = 17.3 L

⇒with n2 : the new number of moles = TO BE DETERMINED

39.2 L / 3.86 moles = 17.3 L / n2

n2 = 1.70 moles

Step 3: Calculate the loss of moles

Loss of moles = 3.86 - 1.70 = 2.16 moles

2.16 moles are lost from the container

Answer 2

$\frac{V_{1} }{n_{1} } =\frac{V_{2} }{n_{2} }$ where V₁ is a volume, n₁ is a number in moles, V₂ is a second volume, and n₂ is its corresponding number in moles.

In our data:

• n₁ = 3.86 mol C₂H₂
• V₁ = 39.2 L
• V₂= 17.3 L
• n₂ = ?

Let's plug our values in the equation above.

$\frac{39.2}{3.86} =\frac{17.3}{n} \\\\10.2=\frac{17.3}{n} \\\\10.2n=17.3\\\\n=1.70$

Note: You could rearrange the equation before plugging in values for less algebraic manipulation.

We're not done yet.  Now, you have to subtract this number of moles from the initial value to find the difference, since the question states "how many moles...have been lost".

$3.86-1.70=2.16$

Answer:

2.16 moles would have been lost