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

56 L of HCl

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.

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

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

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.

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