Balance the following equation stepwise: Ag(s) + HCl(aq) → AgCl + H2

Balanced equation: \(\mathrm{2Ag(s) + 2HCl(aq) \rightarrow 2AgCl\downarrow + H_2\uparrow}\)

Step 1: Count atoms. Left: Ag = 1, H = 1, Cl = 1. Right: Ag = 1, Cl = 1, H = 2.

Step 2: Balance hydrogen. There are 2 H atoms in \(\mathrm{H_2}\), so write \(\mathrm{2HCl}\). Now Cl on the left = 2.

Step 3: Balance chlorine. Write \(\mathrm{2AgCl}\) on the right. Now Ag on the right = 2.

Step 4: Balance silver. Write \(\mathrm{2Ag}\) on the left.

Step 5: Check. Ag: 2 = 2, H: 2 = 2, Cl: 2 = 2. Balanced.

Note: this is a balancing exercise. In practice silver is below hydrogen in the reactivity series, so it does not displace hydrogen from dilute hydrochloric acid.