Question #c0ac4

May 25, 2017

See below.

Explanation:

$\text{Fe(s) + 2H"^"+""(aq)" ⇌ "H"_2"(g)" + "Fe"^"2+""(aq)}$

If the concentration of $\text{H"^"+}$ were increased, the reaction would respond by trying to get rid of the added $\text{H"^"+}$.

The position of equilibrium would shift to the right and the concentration of ${\text{H}}_{2}$ would increase.

The amount of solid does not affect the position of equilibrium.

If some $\text{Fe}$ were removed, there would be no change in the concentration of $\text{Fe"^"2+}$.

If the concentration of ${\text{H}}_{2}$ were increased, the reaction would respond by trying to get rid of the added ${\text{H}}_{2}$.

The position of equilibrium would shift to the left and the mass of $\text{Fe}$ would increase.