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 6.50 M HCl ( aq ) are required to react with 2.55 g Zn ( s ) ?

12.0 milliliters of 6.50 M HCl ( aq ) are required to react with 2.55 g Zn.

Explanation:

moles =

moles of zinc =

The balanced chemical equation is :

According to stoichiometry:

1 mole of zinc reacts with = 2 moles of HCl

Thus 0.0390 moles of zinc reacts with = moles of HCl

To calculate the volume for given molarity, we use the equation:

    .....(1)

Molarity of solution = 6.50 M

Volume of solution = ?

Putting values in equation 1, we get:

Thus 12.0 ml of 6.50 M HCl ( aq ) are required to react with 2.55 g Zn

a gas, having a molecular weight of 13 kg/kmol and a kinematic viscosity of 0.25 cm2/s, flows through a pipe 0.25 m internal diameter and 5 km long at the rate of 0.4 m3/s and is delivered at atmospheric pressure.