a) nZn=m/M=13/65=0,2(mol)
PT:
Zn + 2HCl -> ZnCl2 +H2 \(\uparrow\)
1..........2...........1............1 (mol)
0,2->0,4 -> 0,2 -> 0,2 (mol)
VH2=n.22,4=0,2.22,4=4,48(lít)
b) mHCl=n.M=0,4.36,5=14,6 (g)
=> \(C\%=\dfrac{m_{HCl}.100\%}{m_{ddHCl}}=\dfrac{14,6.100}{500}=2,92\left(\%\right)\)
c)mZnCl2=n.M=0,2.136=27,2(g)
Ta co pthh
Zn + 2HCl \(\rightarrow\) ZnCl2 + H2
Theo de bai ta co
nZn=\(\dfrac{13}{65}=0,2mol\)
a,Theo pthh
nH2=nZn=0,2 mol
\(\Rightarrow VH2_{\left(dktc\right)}=0,2.22,4=4,48l\)
b,Theo pthh
nHCl=2nZn=2.0,2=0,4 mol
\(\Rightarrow mHCl=0,4.36,5=14,6g\)
\(\Rightarrow\) C% cua dd HCl = \(\dfrac{mct}{mdd}.100\%=\dfrac{14,6}{500}.100\%=2,92\%\)
c,Theo pthh
nZnCl2=nZn=0,2 mol
\(\Rightarrow mZnCl2=0,2.127=25,4g\)