Ta có n\(_{H2}=\frac{2,24}{22,4}=0,1mol\)
Zn +2HCl ---->ZnCl2 +H2(1)
0,1>--0,2<-------0,1<-----0,1
ZnO +2HCl--->ZnCl2 +H2O(2)
a)m\(_{Zn}=0,1.65=6,5\left(g\right)\)
m\(_{ZnO}=22,7-6,5=16,2\left(g\right)\)
b)Theo pthh
\(\in\)n\(_{HCl}=2\in n_{H2}=0,2mol\)
C%=\(\frac{0,2.36,5}{600}.100\%=1,22\%\)
pthh: Zn + 2HCl ---> ZnCl2 + H2 (1)
ZnO + 2HCl---> ZnCl2 + H2O
Theo bài ra ta có:
nH2 = 2,24/22,4 = 0,1mol
theo pt (1) ta có:
nZn = nH2 = 0,1mol
--> mZn = 0,1*65 = 6,5(g)
--> mZnO = 22,7 - 6,5 = 16,2g
Zn + 2HCl ---> ZnCl2 + H2 (1)
ZnO + 2HCl ---> ZnCl2 + H2O (2)
nH2 = 2,24/22,4 = 0,1 mol
Theo pt 1, ta có: nZn = nH2 = 0,1 mol
mZn = 0,1.65 = 6,5 g
mZnO = 22,7 - 6,5 = 16,2 g
Theo Pt 1 , ta có: nHCl = 0,1.2 = 0,2 mol
mHCl = 0,2.36,5 = 7,3 g
C%dd HCl = \(\frac{7,3}{600}.100\%\) \(\approx\) 1,22 %