Zinc reacts with hydrochloric acid according to the reaction equation Zn(s)+2HCl(aq)⟶ZnCl2(aq)+H2(g) How many milliliters of 3.50 M HCl(aq) are required to react with 4.95 g Zn(s)?

Answer 1

First, we need to determine the number of moles of Zn:

moles of Zn = mass / molar mass = 4.95 g / 65.38 g/mol = 0.0756 mol

From the balanced chemical equation, we can see that the mole ratio between Zn and HCl is 1:2. This means that 1 mole of Zn reacts with 2 moles of HCl.

So, the number of moles of HCl needed to react with the Zn is:

moles of HCl = 2 x moles of Zn = 2 x 0.0756 mol = 0.1512 mol

Now we can use the formula:

moles = concentration x volume (in liters)

to find the volume of 3.50 M HCl needed to provide 0.1512 moles:

0.1512 mol = 3.50 M x volume (in liters)

volume (in liters) = 0.1512 mol / 3.50 M = 0.0432 L

Finally, we can convert the volume to milliliters:

volume (in milliliters) = 0.0432 L x 1000 mL/L = 43.2 mL

Therefore, 43.2 mL of 3.50 M HCl are required to react with 4.95 g Zn.