Can you calculate the equilibrium concentrations of H2, Br2, & HBr AND the Kc of the reaction

A mixture of 1.374 g of H2 and 70.31 g of Br2 is heated in a 2.00-L vessel at 700 K. These substances act according to:

H2(g) + Br2(g) <------------------> 2HBr(g)

At equilibrium the vessel is found to contain 0.566 g of H2. How do you calculate the equilibrium concentrations of H2, Br2, & HBr? And how do you calculate Kc? Thanks so much for your consideration. Steps please!

H2(g ) + Br2(g) <--------> 2HBr(g)
..1.374 g..70.31 g.............0 g........(start)
.0.566 g ? ?
mole H2 initial = 1374 g H2 x 1 mole / 2.016 g H2 = 0.6815 mole
[H2] = 0.566 g x 1 mol/2.016gH2/2L = 0.141M
mole Br2 remaining = 0.4400 mole - 0.281 mole = 0.159 mole Br2
[Br2] = 0159 mole/ 2.00 L = 0.0795M
mole HBr formed = 2 x (0.6815 - 0.141) = 1.082 mole HBr
? M HBr = 1.082 mol HBr /2.00 L = 0.504

Kc = [HBr}^2/[H2}[Br2] = [0.504]^2 / [0.140]x[0.0795] = 22.7