PTHH: Zn + 2HCl -> ZnCl2 + H2
______0,1___0,2___________0,1
nZn = \(\frac{6,5}{65}\) = 0,1 (mol)
a/ VH2 (đktc) = 0,1*22,4 = 2,24 (lít)
b/ mchất tan HCl = 0,1*36,5 = 3,65 (g)
m = mdung dịch HCl = \(\frac{3,65\cdot100\%}{30\%}\) ≃ 12,17 (g)
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PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
a) Ta có: \(n_{Zn}=\frac{6,5}{65}=0,1\left(mol\right)\) \(\Rightarrow n_{H_2}=0,1mol\)
\(\Rightarrow V_{H_2}=0,1\cdot22,4=2,24\left(l\right)\)
b) Theo PTHH: \(n_{Zn}:n_{HCl}=1:2\) \(\Rightarrow n_{HCl}=0,2mol\)
\(\Rightarrow m_{HCl}=0,2\cdot36,5=7,3\left(g\right)\)
\(\Rightarrow m_{ddHCl}=\frac{7,3}{30\%}\approx24,33\left(g\right)\)
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