Question

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)?

Answer 1

The answer to your question is: 35 ml

Step-by-step 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

Answer 2

34.4 milliliters of 6.00 M HCl

Step-by-step explanation:

Mole = molarity x volume = mass/molar mass,

1 mole of Zn requires 2 moles of HCl

6.75/65.38 mole Zn = 6 x volume of HCl

Therefore,

volume of HCl = 2 x 0.1032/6

= 0.0344 dm3

0.0344 x 1000 = 34.4 milliliters.

Hence, 34.4 milliliters of HCl will be required to react with 6.75 g Zn.