Question

How many milliliters of 3.00 M HCl(aq) are required to react with 4.15 g of Zn(s)?

Answer 1

Balanced equation is

2 HCl + Zn ------ > ZnCl2 + H2

Number of mole sof Zn = 4.15 g / 65.38 g/mol = 0.0635 mole

From the balanced equation we can say that

1 mole of Zn requires 2 mole of HCl so

0.0635 mole of Zn will require

= 0.0635 mole of Zn *(2 mole of HCl / 1 mole of Zn)

= 0.127 mole of HCl

Molarity of HCl = number of moles of HCl / volume of solution inL

3.00 = 0.127 / volume of solution in L

volume of solution in L = 0.127 / 3.00 = 0.0423 L

1L = 1000 mL

0.0423 L = 42.3 mL

Therefore, the volume of HCl required would be 42.3 milliliter