Question

consider this reaction:
h₂ (g) + cl₂ (g) ⟶ 2hcl (g)
how many liters of hcl are produced when 11.2 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

○ 112 l

○ 56 l

○ 224 l

Explanation:

Step1: Calculate moles of \(Cl_2\)

Given \(V_{Cl_2}=11.2\ L\). Using \(n=\frac{V}{V_m}\) (where \(V_m = 22.4\ L/mol\)), \(n_{Cl_2}=\frac{11.2\ L}{22.4\ L/mol}=0.5\ mol\)

Step2: Use mole ratio from reaction

From \(H_2(g)+Cl_2(g)\to2HCl(g)\), mole ratio \(Cl_2:HCl = 1:2\). So \(n_{HCl}=2\times n_{Cl_2}=2\times0.5\ mol = 1\ mol\)

Step3: Calculate volume of \(HCl\)

Using \(V = n\times V_m\), \(V_{HCl}=1\ mol\times22.4\ L/mol = 22.4\ L\)

Answer:

22.4 L