How to calculate the rate of consumption of aluminium with hydrochloric acid?
Since one of the products is a gas $\ce{H2}$, the decrease in system mass must reflect the escape of hydrogen. Thus, after thirty seconds, the system has lost $\pu{0.030 g}$ of hydrogen, which is $\pu{0.01488 mol}$ of $\ce{H2}$. 

Thus, the rate $r$ of hydrogen production is:

$$
r (\ce{H2})
= \frac{\Delta n(\ce{H2})}{\Delta t}
= \frac{\pu{0.01488 mol}}{\pu{30 s}}
= \pu{4.96E-4 mol//s}$$

Now, how can you use this rate and the stoichiometry of the reaction to determine the rate of aluminum consumption?


   $\ce{2 Al (s) + 6 HCl (aq) -&gt; 3 H2 (g) + 2 AlCl3 (aq)}$
 For every three moles of hydrogen produced two moles of aluminium are consumed, therefore $r(\ce{Al}) = -\frac{2}{3}r(\ce{H2}) = \pu{3.31E-4 mol//s}$.