Zn +2HCl --->ZnCl2 +H2
a) Ta có
n\(_{Zn}=\frac{13}{65}=0,2mol\)
Theo pthh
n\(_{H2}=n_{Zn}=0,2mol\)=> m\(_{H2}=0,2.2=0,4\left(g\right)\)
V\(_{H2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\)
b)Theo pthh
n\(_{ZnCl2}=n_{Zn}=0,2mol\)
=>m\(_{ZnC_{ }l2}=0,2.136=27,2\left(g\right)\)
Theo pthh
n\(_{HCl}=2n_{Zn}=0,4mol\)
=>m\(_{HCl}=0,4.36,5=14,6\left(g\right)\)
m\(_{dd}\)sau pư=100+14,6-0,4=114,2(g)
C%=\(\frac{27,2}{114,2}.100\%=23.82\%\)
Chúc bạn hok tốt
có: nZn= \(\frac{13}{65}\)= 0,2( mol)
PTPU
Zn+ 2HCl\(\rightarrow\) ZnCl2+ H2\(\uparrow\)
.0,2................0,2........0,2...... mol
\(\Rightarrow\) VH2= 0,2. 22,4= 4,48( lít)
có: mZnCl2= 0,2. 136= 27,2( g)
mdd sau pư= 13+ 100- 0,2.2
= 112,6( g)
\(\Rightarrow\) C%ZnCl2= \(\frac{27,2}{112,6}\). 100%= 24,16%
