CHEMICAL EQUILIBRIUM!! Consider the following reversible reaction. In a 3.00-liter container, the following amounts are found in equilibrium at 400°C: 0.0420 mole N2, 0.516 mole H2, and 0.0357 mole NH3. Evaluate Kc . N2(g) + 3H2(g) 2NH3(g) a. 0.202 b. 1.99 c. 16.0 d. 4.94 e. 0.503 Answer is B I understand the setup but what do I do with the 3.00L and 400 degrees C? How could I get the answer b??

Question

CHEMICAL EQUILIBRIUM!!
Consider the following reversible reaction. In a 3.00-liter container, the following amounts are found in equilibrium at 400°C: 0.0420 mole N2, 0.516 mole H2, and 0.0357 mole NH3. Evaluate 
Kc
. N2(g) + 3H2(g) <---> 2NH3(g)
a. 0.202 b. 1.99 c. 16.0 d. 4.94 e. 0.503

Answer is B
I understand the setup but what do I do with the 3.00L and 400 degrees C? How could I get the answer b??