Дано
m(Cu)=6.4 g
m(ppaHNO3)=126 g
W(HNO3)=20\%
η=90\%
--------------------------
V(NO2)-?
m(HNO3)=126*20\%/100\%=25.2 g
M(Cu)=64 g/mol
n(Cu)=m/M=6.4 / 64=0.1 mol
M(HNO3)=63 g/mol
n(HNO3)=m/M=25.2 / 63=0.4 mol
n(Cu)<n(HNO3)
6,4 X
Cu+4HNO3(p)-->Cu(NO3)2+2NO2+2H2O Vm=22.4 L/mol
64 2*22.4
X=6.4 * 44.8 / 64 = 4.48 L
V(NO2)=4.48*90\% / 100\%
V(NO2)=4.032 L
ответ 4.032 л
Na+2HOH→ 2NaOH +H2
28r. xr
28r/23r=xr/80r=>
xr=(28r*80r)/23r
xr=97r.
V(NaOH)=97/40=2,425моль.