How to calculate the rate of consumption of aluminium with hydrochloric acid? [closed]








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> 
> An experiment is done to determine the rate of the following reaction:
>  $$\ce{2 Al (s) + 6 HCl (aq) -> 3 H2 (g) + 2 AlCl3 (aq)}$$
> 
> 
> The following data is collected: 
>  \begin{array}{rr} 
> \text{time}/\pu{s} & (\text{Mass of beaker + contents})/\pu{g}\\\hline
>  0.0 & 270.230\\
>  30.0 & 270.200\\
>  60.0 & 270.170 \end{array}
> 
> 
> Calculate the rate of consumption of $\ce{Al}$ in $\pu{g/min}$.
> 
> 
> 


The answer is $\pu{0.534 g/min}$, but I don't know how to get there.




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?



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


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