Question

Oxidation of zinc produced a `47.5*mL` volume of dihydrogen at `1*atm` and `273.15*K`. What is the molar quantity of zinc that reacted?

Answer 1

I will do this for `1*atm`, and `273.15*K`.........

Explanation:

We interrogate the redox reaction...........

`Zn(s) + 2HCl(aq) rarr ZnCl_2(aq) + H_2(g)`

We use the old Ideal gas equation to access the molar quantity....

`n=(PV)/(RT)=(1*atmxx47.5*mLxx10^-3*L*mL^-1)/(0.0821*L*atm*K^-1*mol^-1xx273.15*K)`

`=2.12xx10^-3*mol`.

I am reluctant to use `"SATP"` because the values seem to change across curricula. You will have to adapt this question to whatever values of `"SATP"` apply. As a chemist, I tend to like measurement of pressure in `"atmospheres"`, and measurement of volume in `"millilitres"`. As a student you have to be adaptable.