nZn = 6.5/65=0,1mol
pt : Zn + HCl -------> ZnCl2 + H2
npứ: 0,1------------------>0,1------>0,1
VH2 = 0,1.22,4=2,24l
mZnCl2=0,1.136=13,6g
mH2 = 0,1.2=0,2g
mdd ZnCl2 = m Zn + mddHCl - mH2
= 6,5 +200-0,2= 206,3g
\(C\%\left(ZnCl_2\right)=\dfrac{13,6}{206,3}.100\approx6,59\%\)
nZn=6,5/65=0,1 mol
Zn +2HCl----->ZnCl2+H2
0,1--->0,2---->0,1----->0,1
a, VH2=0,1x 22,4=2,24(l)
B, mZnCl2=0,1*136=13,6(g)
mH2=0,1*2=0,2(g)
mdd=200+6,5-0,2=206,3(g)
C%=13,6/206,3*100%~6,6%
a) Zn + 2HCl → ZnCl2 + H2↑
b) \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
Theo PT: \(n_{H_2}=n_{Zn}=0,1\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,1\times22,4=2,24\left(l\right)\)
c) Theo PT: \(n_{HCl}=2n_{Zn}=2\times0,1=0,2\left(mol\right)\)
\(\Rightarrow m_{HCl}=0,2\times36,5=7,3\left(g\right)\)
\(\Rightarrow C\%_{HCl}=\dfrac{7,3}{200}\times100=3,65\%\)
