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#
