Zn(s)+2HCl(aq) ->ZnCl2(aq)+Hx(g)

How many millilters of 2.50M HCl(aq) are required to react with 3.95 of Zn(s)?

mols Zn = grams/atomic mass = ?

mols HCl = twice that (from the coefficients in the balanced equation).
M HCl = mols HCl/L HCl. You know M and mols, solve for L HCl and convert to mL.

To solve this problem, we need to use the concept of stoichiometry.

Step 1: Write and balance the equation:
Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)

Step 2: Determine the molar mass of Zn:
The molar mass of Zn is 65.38 g/mol.

Step 3: Convert the mass of Zn to moles:
3.95 g Zn * (1 mol Zn / 65.38 g Zn) = 0.0603 mol Zn

Step 4: Use the balanced equation to set up the stoichiometric ratio:
From the balanced equation, we can see that the ratio of moles of Zn to moles of HCl is 1:2.
Therefore, 0.0603 mol Zn * (2 mol HCl / 1 mol Zn) = 0.1206 mol HCl

Step 5: Convert moles of HCl to milliliters of HCl solution:
To convert moles to milliliters, we need to use the molarity (M) of the HCl solution.
Molarity (M) = moles / liters

Let's calculate the volume of 2.50 M HCl solution required:
Molarity (M) = 2.50 mol/L

0.1206 mol HCl * (1 L / 2.50 mol) * (1000 mL / 1 L) = 48.24 mL

Therefore, 48.24 milliliters of 2.50 M HCl solution are required to react with 3.95 grams of Zn.