How many milliliters of 3.50 m hcl(aq) are required to react with 5.25 g of zn(s)?

Reaction: 2HCl + Zn → ZnCl₂ + H₂

Now, Molar mass of Zn = Given mass / Molecular mass
Molar mass = 5.25 / 65.38  [ Molar mass of Zn = 65.38 g/mol ] 
Molar mass = 0.0802 mol

Now, As reaction requires twice as many moles of HCl as it does Zn, we will need = 0.0802 × 2 = 0.1605 moles of HCl

We know, Volume = moles / molarity 
Here, moles = 0.1605 mol
molarity = 3.50 M

Substitute their values in the formula, 
Volume = 0.1605 / 3.50
V = 0.04588 L
V = 45.88 mL

So, final answer of your question would be 45.88 mL

Hope this helps!

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