For the reaction;
h2(g) + br2 ⇌ 2hbr (g)
if i start with 0.10 m hydrogen and 0.20 m bromine what are the equilibrium concentrations of each if kc = 62.5?

At equilibrium:

[H2] = 0.005 M

[Br2] = 0.105 M

[HBr] = 0.189 M

Explanation:

H2(g) + Br2(g) ⇄ 2HBr

an "x" value will be used from reactant to produced "2x"

so at equilibrium:

[H2] = 0.1 - x

[Br2] = 0.2 - x

[HBr] = 2x

we know that Kc=[HBr]²/[H2][Br2]

Thus 62.5 = (2x)²/(0.1-x)(0.2-x)

this generate a quadratic equation: 58.5x² - 18.75x + 1.25 = 0

the x₁ = 0.23   x₂ = 0.09457

we pick 0.09457 because the two reactants can not make more than what they have. x₁ is higher than both initial reactant concentration

Then we substitute the "x₂" value at equilibrium:

[H2] = 0.1-0.09457 = 0.005 M

[Br2] = 0.2-0.09457 = 0.105 M

[HBr] = 2*0.09457 = 0.189 M

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i think it is true since you can test the answer and can use observations to complete it.