Post a New Question

Chemisty

posted by .

A 0.411 g sample of powdered copper mixed with an inert, soluble additive was fully consumed by 23.4mL of 0.602M nitric acid, producing copper (II) nitrate, water, and nitric oxide. What is the percent copper by mass in the sample?

3Cu + 8HNO3 = 3Cu(NO3)2 +2NO =4H2O

  • Chemisty -

    Dear Veronica,
    number of moles of HNO3 used=
    0.602x(23.4/1000)
    = 0.014
    number of mol of Cu used=
    0.014/8x3
    =5.28255x10^-3

    mass of the Cu involved in the reaction
    = 5.28255x10^-3 x 63.5
    = 0.33544g

    %by mass of Cu =
    0.33544/0.411 x 100%
    = 81.6%

Answer This Question

First Name
School Subject
Your Answer

Related Questions

More Related Questions

Post a New Question