Question

A popular classroom demonstration consists of filing notches into a new penny and soaking the penny in hydrochloric acid overnight. Because new pennies are made of zinc coated with copper, and hydrochloric acid dissolves zinc and not copper, the inside of the penny is dissolved by the acid, while the outer copper shell remains. Suppose a penny contains 2.5 g of zinc and is soaked in 20.0 mL of 6.0 M HCl. Calculate the concentra- tion of the HCl solution after all of the zinc has dissolved. Hint: The Zn from the penny is oxidized to Zn2

Answer 1

`HCl conc=2.2M`

Step-by-step explanation:

From the question we are told that:

Mass of zinc
`Mass_(z_n)=2.5g`

Initial conc of HCl
`HConc_1=6.0M`

Initial volume of HCl
`Hvol_1=20.0mL`

Molar mass of zinc
`M_(Z_n)=65.38g/mol`

Generally the equation for reaction is mathematically given by

`2HCL+Zn\implies ZnCl_2 +H_2`

Generally the equation for moles of zinc
`m_(z_n)` is mathematically given by

`m_(z_n)=(Mass_(Z_n))/(M_(Z_n))`

`m_(z_n)=(2.5g)/(M_(65.38))`

`m_(z_n)=0.038molZn`

Generally the equation for moles of
`m_(HCl)` is mathematically given by

`m_(HCl)=HConc_1*Hvol_1`

`m_(HCl)=6.0M*20.0mL`

`m_(HCl)=6.0M*20.0*10^(-3)L`

`m_(HCl)=0.12\ mol\ HCl`

Generally the reacted moles of HCl
`HCl_[reacted]` is mathematically given by

Since

Zn:HCl =1:2

Therefore

`HCl_[reacted]=0.038*(2)/(1)`

`HCl_[reacted]=0.076mol \ HCl`

Generally the moles of HCl after Zn oxidization X is mathematically given by

`x=0.12-0.076`

`x=0.044 mol HCl`

Generally the conc of hydrochloride acid X is mathematically given by

`X=(0.044)/(0.020)`

`X=2.2 M Hcl`

Therefore Conc of HCl

`HCl conc=2.2M`