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Answer 1 · 2017-05-25T00:22:56

See below.

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

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

The position of equilibrium would shift to the right and the concentration of $H_{2}$ $"H"_2$ would increase.

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

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

If the concentration of $H_{2}$ $"H"_2$ were increased, the reaction would respond by trying to get rid of the added $H_{2}$ $"H"_2$ .

The position of equilibrium would shift to the left and the mass of $Fe$ $"Fe"$ would increase.