Question

H2(g)+Cl2→2HCl(g)

How many liters of HCl are produced when 28 L of chlorine are reacted with excess hydrogen?

(One mole of any gas occupies 22.4 L under certain conditions of temperature and pressure. Assume those conditions for this question.)


22.4 L


56 L


112 L


224 L

Answer 1

56L

Step-by-step explanation:

Got it right on the test.

Answer 2

56 L of HCl

Step-by-step explanation:

The equation for the reaction is;

H₂(g )+ Cl₂(g) →2HCl(g)

Volume of Chlorine gas = 28 L

According to the question, 1 mole of a gas occupies 22.4 L at stp.

We are required to determine the volume of HCl produced.

Step 1: Moles of Chlorine gas that reacted

Volume of Chlorine gas = 28 L

But;

1 mole = 22.4 L

Therefore;

Moles of Chlorine gas = 28 L ÷ 22.4 L

= 1.25 moles of Cl₂

Step 2: Number of moles of HCl

From the equation 1 mole of Cl₂ reacts with hydrogen gas to produce 2 moles of HCl

Therefore, the mole ratio of Cl₂ to HCl is 1: 2

Thus; moles of HCl = 1.25 × 2

= 2.5 moles

Step 3: Volume of HCl gas prdouced

We know that. 1 mole of a gas = 22.4 L

Thus,

The volume of HCl = 2.5 moles × 22.4 L/mol

=56 L of HCl

Therefore, 56 L of HCl are produced when 28 L of chlorine gas reacts with excess hydrogen gas.