Calculate the amount of hydrogen gas produced when 1.96 g of zinc reacts with 100 cm3 of 0.2 M HCl.

The molar mass of zinc ($\text{Zn}$) is 65.4 g/mol.

$$\text{Moles of Zn} = \frac{\text{Mass of Zn}}{\text{Molar mass of Zn}}$$ $$\text{Moles of Zn} = \frac{1.96 \text{ g}}{65.4 \text{ g/mol}}$$ $$\text{Moles of Zn} \approx 0.02997 \text{ mol}$$

The volume of $\text{HCl}$ is 100 cm$^3$, which is 0.100 L. The concentration is 0.2 M.

$$\text{Moles of HCl} = \text{Concentration} \times \text{Volume (in L)}$$ $$\text{Moles of HCl} = 0.2 \text{ mol/L} \times 0.100 \text{ L}$$ $$\text{Moles of HCl} = 0.020 \text{ mol}$$

The balanced chemical equation is:

$$\text{Zn(s)} + 2\text{HCl(aq)} \rightarrow \text{ZnCl}_2\text{(aq)} + \text{H}_2\text{(g)}$$

From the equation, 1 mole of $\text{Zn}$ reacts with 2 moles of $\text{HCl}$.

Let's determine how much $\text{HCl}$ is needed to react with the available $\text{Zn}$:

$$\text{Moles of HCl needed} = \text{Moles of Zn} \times \frac{2 \text{ mol HCl}}{1 \text{ mol Zn}}$$ $$\text{Moles of HCl needed} = 0.02997 \text{ mol Zn} \times 2$$ $$\text{Moles of HCl needed} \approx 0.05994 \text{ mol HCl}$$

We have 0.020 mol of $\text{HCl}$, which is less than the 0.05994 mol needed. Therefore, $\text{HCl}$ is the limiting reagent, and $\text{Zn}$ is in excess.

Alternatively, let's determine how much $\text{Zn}$ is needed to react with the available $\text{HCl}$:

$$\text{Moles of Zn needed} = \text{Moles of HCl} \times \frac{1 \text{ mol Zn}}{2 \text{ mol HCl}}$$ $$\text{Moles of Zn needed} = 0.020 \text{ mol HCl} \times \frac{1}{2}$$ $$\text{Moles of Zn needed} = 0.010 \text{ mol Zn}$$

We have 0.02997 mol of $\text{Zn}$, which is more than the 0.010 mol needed. Therefore, $\text{Zn}$ is in excess.

The reagent that was in excess is $\text{Zinc}$.