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?

1 Answer
Jun 14, 2017

Answer:

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.