# Question #72a23

Jan 30, 2017

Because in (B) you're only changing the volume of the acid, and it was already in excess.

#### Explanation:

The reaction is: $Z n \left(s\right) + 2 H C l \left(a q\right) - \to Z n C {l}_{2} \left(a q\right) + {H}_{2} \left(g\right)$

You need 2 moles of HCl to react with each mole of Zinc. Zinc has a molar mass of 65 g/mol so 1 g of zinc is 0.015 moles.

However, in 100 $c {m}^{3}$ of 1M HCl you would have 0.1 moles of HCl. You need 2 x 0.015 = 0.03 moles of HCl, but you have 0.1 moles - in other words 3 times as much as you need. So the HCl is already well in excess.

If you increase the volume of the HCl to 200 $c {m}^{3}$ you would simply have an even greater excess. So the net result on the rate of reaction would be: Nothing. You would just have more unreacted HCl left over at the end.

To change the rate of the reaction you need to increase the ease of collision between the ${H}^{+}$ ions and the zinc atoms. So you could do this by increasing acid concentration (more ions per unit volume, so more collisions), or increase the temperature (more mobility for ions, so more collisions) or increase surface area of the zinc particles (greater area on which ions can react). But adding even more of something you already had too much of.....no difference.