How much additional hydrochloric acid should be added to arrive at ‘excess acid’?
The volume you've established seems correct, as well as your assumption that extra $\pu{40 mL}$ of $\ce{HCl}$ would be a good excess:

$$\ce{NaHCO3 + HCl -&gt; NaCl + H2O + CO2}$$

$$V(\ce{HCl}) = \frac{n(\ce{HCl})}{C(\ce{HCl})} = \frac{m(\ce{NaHCO3})}{M(\ce{NaHCO3}) \cdot C(\ce{HCl})} = \frac{\pu{0.5 g}}{\pu{84.0 g mol-1} \cdot \pu{0.1 mol L-1}} = \pu{59.5 mL}$$

I suspect your teacher either prepared a series of various antacids, or this one contains additional components, such as $\ce{CaCO3}$, $\ce{Mg(OH)2}$ of $\ce{Al(OH)3}$. Note that for the neutralization of the same mass of, say, aluminium(III) hydroxide, over than 3 times more $\ce{HCl}$ is required:

$$\ce{Al(OH)3 + 3 HCl -&gt; AlCl3 + 3 H2O}$$

$$V(\ce{HCl}) = \frac{n(\ce{HCl})}{C(\ce{HCl})} = \frac{3m(\ce{Al(OH)3})}{M(\ce{Al(OH)3}) \cdot C(\ce{HCl})} = \frac{3 \cdot \pu{0.5 g}}{\pu{78.0 g mol-1} \cdot \pu{0.1 mol L-1}} = \pu{192 mL}$$

Either way, it looks like you understand how to properly find out the amount of acid for the neutralization. I think if you demonstrate your detailed attempt to the teacher, it should clarify the situation.