Zinc reacts with hydrochloric acid according to the reaction equation Zn (s) + 2 HCl (aq) ⟶ ZnCl 2 (aq) + H 2 (g) How many milliliters of 5.00 M HCl (aq) are required to react with 3.15 g Zn (s) ?

We need 19.3 mL of HCl

Explanation:

Step 1: Data given

Molarity HCl = 5.00 M

Mass Zn = 3.15 grams

Step 2: The balanced equation

Zn (s) + 2HCl (aq) ⟶ ZnCl2 (aq) + H2 (g)

Step 3: Calculate moles Zn

Moles Zn = mass Zn / molar mass Zn

Moles Zn = 3.15 grams / 65.38 g/mol

Moles Zn = 0.0482 moles

Step 4: Calculate moles HCl needed

For 2 moles HCl we need 1 mol Zn to produce 1 mol ZnCl2 and 1 mol H2

For 0.0482 moles Zn we need 2*0.0482 = 0.0964 moles HCl needed

Step 5: Calculate volume needed

Molarity = moles / volumes

Volumes = moles / molarity

Volume HCl needed = 0.0964 moles HCl / 5.00 M

Volume HCl needed = 0.01928 L = 19.3 mL

We need 19.3 mL of HCl