Write the equilibrium expression for this reaction: (c19h8br4o5s)h– ↔ (c19h8br4o5s)2– + h+

Explanation:

The equilibrium expression for the general equilibrium represented by the equation:

is:

Then, following that sketch, you obtain:

        (C₁₉ H₈ Br₄ O₅ S) H⁻ ⇄ (C₁₉ H₈ Br₄ O₅ S)²⁻ + H⁺

         Keq = [ (C₁₉ H₈ Br₄ O₅ S)²⁻ ] [ H⁺ ] / [ (C₁₉ H₈ Br₄ O₅ S) H⁻  ]

The square brackets are used to mean molar concentration (molariy).

the answer to your question is c

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2