Question

A reaction occurs in a mixture of 2 liters of hydrogen gas and 3 liters of chlorine gas. How many liters of hydrochloric gas are produced? Which is the excess and by how much?

Answer 1

We will produce 4.0 L of hydrochloric gas

Chlorine gas is the excess. There will remain 1.0 L

Step-by-step explanation:

Step 1: Data given

Volume o H2 = 2.0 L

Volume of Cl2 = 3.0 L

Step 2: The balanced equation

H2 + Cl2 → 2HCl

Step 3: Calculate moles

22.4 = 1 mol

2.0L H2 = 0.0893 moles H2

3.0 L Cl2 = 0.134 moles Cl2

Step 4: Calculate limiting reactant

For 1 mol H2 we need 2 moles Cl2 to produce 2 moles HCl

H2 has the smallest amount of moles so it's the limiting reactant.

It will completely be consumed (0.0893 moles)

Cl2 is in excess. There reacts 0.0893 moles

There will remain 0.134 - 0.0893 = 0.0447 moles

This is 0.0447 * 22.4 = 1.0 L

Step 5: Calculate moles HCl

For 1 mol H2 we need 2 moles Cl2 to produce 2 moles HCl

For 0.0893 moles H2 we'll have 2*0.0893 = 0.1786 moles HCl

Step 6: Calculate volume HCl

1 mol = 22.4 L

0.1786 moles = 4 L

We will produce 4.0 L of hydrochloric gas

Chlorine gas is the excess. There will remain 1.0 L