Titanium and chlorine react to form titanium(IV) chloride, like this: Ti(s)+2Cl2(g)→TiCl4(l) At a certain temperature, a chemist finds that a 6.3L reaction vessel containing a mixture of titanium, chlorine, and titanium(IV) chloride at equilibrium has the following composition: compound amount Ti 2.97g Cl2 2.42g TiCl4 3.32g Calculate the value of the equilibrium constant Kc for this reaction. Round your answer to 2 significant digits. g

Kc = 3.4 × 10⁴

Explanation:

Let's consider the following reaction at equilibrium.

Ti(s) + 2 Cl₂(g) → TiCl₄(l)

The concentration equilibrium constant (Kc) includes only gases and aqueous species. So, in this case, we have to calculate the molarity of Cl₂(g) using the following expression.

M = mass of solute / molar mass of solute × liters of solution

[Cl₂] = 2.42 g / 70.91 g/mol × 6.3 L

[Cl₂] = 5.42 × 10⁻³ M

The concentration equilibrium constant (Kc) is:

Kc = 1 / [Cl₂]²

Kc = 1 / (5.42 × 10⁻³)²

Kc = 3.4 × 10⁴

i think so it’s will be b yes i believe its

it would be b. 6co2 + 6h2o --> c6h12o6 + 6o2