Bạn sửa lại đề là : 36.5 % nhé
nH2 = 2.24/22.4 = 0.1 mol
Zn + 2HCl --> ZnCl2 + H2
0.1___0.2______0.1___0.1
mZn = 6.5 g
mZnO = 14.6 - 6.5 = 8.1g
nZnO = 0.1 mol
ZnO + 2HCl --> ZnCl2 + H2O
0.1____0.2______0.1
%Zn = 44.52%
%ZnO = 55.48%
nZnCl2 = 0.2 mol
mZnCl2 = 27.2 g
nHCl = 0.4 mol
mHCl = 0.4*36.5 = 14.6g
mddHCl = 14.6*100/36.5 = 40g
mdd sau phản ứng = 14.6 + 40 - 0.2 = 54.4 g
C%ZnCl2 = 27.2/54.4*100% = 50%
Zn + 2HCl → ZnCl2 + H2 (1)
ZnO + 2HCl → ZnCl2 + H2O (2)
\(n_{H_2}=\frac{2,24}{22,4}=0,1\left(mol\right)\)
a) Theo Pt1: \(n_{Zn}=n_{H_2}=0,1\left(mol\right)\)
\(\Rightarrow m_{Zn}=0,1\times65=6,5\left(g\right)\)
\(\%m_{Zn}=\frac{6,5}{14,6}\times100\%=44,52\%\)
\(\%m_{ZnO}=100\%-44,52\%=55,48\%\)
b) Theo PT1: \(n_{HCl}=2n_{Zn}=2\times0,1=0,2\left(mol\right)\)
\(m_{ZnO}=14,6-6,5=8,1\left(g\right)\)
\(\Rightarrow n_{ZnO}=\frac{8,1}{81}=0,1\left(mol\right)\)
Theo PT2: \(n_{HCl}=2n_{ZnO}=2\times0,1=0,2\left(mol\right)\)
\(\Rightarrow\Sigma n_{HCl}=0,2+0,2=0,4\left(mol\right)\)
\(\Rightarrow\Sigma m_{HCl}=0,4\times36,5=14,6\left(g\right)\)
\(\Rightarrow m_{ddHCl}=\frac{14,6}{36\%}=40,56\left(g\right)\)
c) \(m_{H_2}=0,1\times2=0,2\left(g\right)\)
Ta có: \(m_{dd}saupứ=14,6+40,56-0,2=54,96\left(g\right)\)
Theo Pt1: \(n_{ZnCl_2}=n_{Zn}=0,1\left(mol\right)\)
Theo PT2: \(n_{ZnCl_2}=n_{ZnO}=0,1\left(mol\right)\)
\(\Rightarrow\Sigma n_{ZnCl_2}=0,1+0,1=0,2\left(mol\right)\)
\(\Rightarrow\Sigma m_{ZnCl_2}=0,2\times136=27,2\left(g\right)\)
\(\Rightarrow C\%_{ZnSO_4}=\frac{27,2}{54,96}\times100\%=49,49\%\)
