Zinc reacts with hydrochloric acid according to the reaction equation zn(s)+2hcl(aq)⟶zncl2(aq)+h2(g)

how many milliliters of 6.00 m hcl(aq) are required to react with 6.75 g zn(s)?

The answer to your question is:  35 ml

Explanation:

Data

Volume of HCl = ? concentration = 6.0 M

mass of Zinc = 6.75 g

Equation

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

MW Zn = 65.4 g

MW HCl = 2 x (35.5 + 1) = 73 g

                      65.4 g of Zn     73 g of HCl

                     6.75 g of Zn          x

                     x = (6.75 x 73) / 65.4

                    x = 7.53 g of HCl

                     36.5 g of HCl  1 mol

                     7.53 g of HCl  x

                     x = (7.53 x 1)/36.5

                    x = 0.21 mol of HCl

Molarity = moles / volume

Volume = moles / molarity

Volume = 0.21 / 6

Volume = 0.035 l or 35 ml