Zn +2HCl---> ZnCl2 +H2
Ta có
n\(_{Zn}=\frac{32,5}{65}=0.5mol\)
n\(_{HCl}=\frac{43,8}{36,5}=1,2mol\)
Lập tỉ lệ
n\(_{Zn}=\frac{0,5}{1}=0,5mol\)
n\(_{HCl}=\frac{1,2}{2}=0,6\)
=>HCl dư
vậy chất sau pư gồm HCl dư, ZnCl2, H2
theo pthh
n\(_{HCl}=2n_{Zn}=0=1mol\)
=>n\(_{HCl}dư=1,2-1=0,2mol\)
m\(_{HCl}dư=0,2.36,5=7,3\left(g\right)\)
Theo pthh
n\(_{ZnCl2}=n_{Zn}=0,5mol\)
m\(_{ZnCl2}=0,5.136=68\left(g\right)\)
Theo pthh
n\(_{H2}=n_{Zn}=0,5mol\)
m\(_{H2}=0,5.2=1\left(g\right)\)
Do H% =80 suy ra
m\(_{HCl}dư=\frac{7,3.80}{100}=5,84\left(g\right)\)
m\(_{ZnCl2}=\)\(\frac{68.80}{100}=54,4\left(g\right)\)
m\(_{H2}=\frac{1.80}{100}=0,8\left(g\right)\)
nZn = 32,5/65 = 0,5 mol
Zn +2HCl --> ZnCl2 + H2
0,5 ---------> 0,4 ---> 0,4
=> mZnCl2 = 0,4.136= 54,4 g
vH2 = 0,4.22,4 = 8,96 lít
Tham khảo
nZn = 32.5/65 = 0.5 mol
nHCl = 1.2 mol
Zn + 2HCl --> ZnCl2 + H2
Bđ: 0.5___1.2
Pư: 0.5____1________0.5___0.5
Kt: 0 _____0.2_______0.5___0.5
Vì : H% = 80%
nHCl dư = 0.16 mol
nZnCl2 = 0.4 mol
nH2 = 0.4 mol
mHCl dư = 5.84 g
mZnCl2 = 54.4 g
mH2 = 0.8 g
