Given mass of Cu = 18.9 g
Molar mass of Cu 63.55 g/mol
No. of moles of Cu = 18.9/63.55 = 0.297 moles ............(1)
Given Conentration of Nitric acid = 16 mol/L
Volume = 82 ml = 0.082L
Molarity = n/V
where n = no. of moles of HNO3 and V = volume in L
n = 16/0.082 = 1.312 moles .................(2)
Chemical equation
Cu + 4HNO_3 --> Cu(NO_3)_2 + 2H_2O + 2NO_2
According to equation
1 mole of Cu produces 2 moles of 2NO_2
so, 0.297 moles of Cu will produce 2NO_2 = 2 X 0.297 = 0.594 moles...................(3)
In equation
4 moles of 4HNO_3 produces 2 moles of 2NO_2
so, 1.312 moles of 4HNO_3 will produce 2NO_2 = (2 x 1.312)/ 4 = 0.656 moles ............(4)
As given moles of Cu produces lesser moles of 2NO_2 hence Cu is the limiting reagent
a. The limiting reagent in the reaction is Copper
b. As solved above
0.297 moles of Cu will produce 2NO_2 = 2 X 0.297 = 0.594 moles
2NO_2 produced = 0.594 moles
Molar mass of 2NO_2 = 46 g/mol
mass = 0.594 moles X 46 g/mol = 27.324g
2NO_2 = 27.324g
