Respostas

2014-03-16T12:38:58-03:00
  [H2] = 0,5 / 5 = 0,1

[ I2] = 0,3 / 5 = 0,06

     H2(g) + I2(g) ~>  2HI(g)

Kc =  \frac{[HI]^2}{[H2] . [I2]}

 46 = \frac{[HI]^2}{[0,1] . [0,06]}

[HI] =  \sqrt{0,276} =0 ,525 mol / L

0,525 mol ------1L
x -------------------5L

x =2,625 mol