The equilibrium constant, k of the reaction in which case, the concentrations of the given reactants and products are as indicated is; Choice A; K = 3.1 x 10⁵
<h3>What is the equilibrium constant , k of the reaction as described in the task content?</h3>
It follows from above that the concentrations of the reactants and products are as follows; [H2] = 0.10 M, [N2] = 0.10 M, and [NH3] = 5.6 M at equilibrium.
Hence, the equilibrium constant of the reaction in discuss is;
K = [5.6]²/[0.10]³[0.10]
k = 5.6² × 10⁴
k = 3.136 × 10⁵
K = 3.1 × 10⁵.
Read more on equilibrium constant;
brainly.com/question/1619133
#SPJ1
<u>Answer:</u> 6.57 L of solution can be made.
<u>Explanation:</u>
Molarity is defined as the amount of solute expressed in the number of moles present per liter of solution. The units of molarity are mol/L. The formula used to calculate molarity:
.....(1)
Given values:
Molarity of LiBr = 3.5 M
Moles of LiBr = 23 moles
Putting values in equation 1, we get:

Hence, 6.57 L of solution can be made.
Answer:
Limiting reactants or limiting reagents decide the amount of product formed and the amount of excess reagent used.
3 NaOH + FeBr3 = 3 NaBr + Fe(OH)3
Hope this helps!!