PTHH: Zn + H2SO4 → ZnSO4 + H2
\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
a) Theo PT: \(n_{Zn}pư=n_{H_2}=0,3\left(mol\right)\)
\(\Rightarrow m_{Zn}pư=0,3\times65=19,5\left(g\right)\)
b) \(m_{H_2SO_4}=400\times12,25\%=49\left(g\right)\)
\(\Rightarrow n_{H_2SO_4}=\dfrac{49}{98}=0,5\left(mol\right)\)
Theo PT: \(n_{H_2SO_4}=n_{H_2}\)
Theo bài: \(n_{H_2SO_4}=0,5>n_{H_2}=0,3\)
⇒ dd H2SO4 dư
Vậy dung dịch sau phản ứng gồm: H2SO4 dư và ZnSO4
Theo PT: \(n_{H_2SO_4}pư=n_{H_2}=0,3\left(mol\right)\)
\(\Rightarrow n_{H_2SO_4}dư=0,5-0,3=0,2\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4}dư=0,2\times98=19,6\left(g\right)\)
Theo PT: \(n_{ZnSO_4}=n_{H_2}=0,3\left(mol\right)\)
\(\Rightarrow m_{ZnSO_4}=0,3\times161=48,3\left(g\right)\)
\(\Sigma m_{dd}=19,5+400=419,5\left(g\right)\)
\(\Rightarrow C\%_{H_2SO_4}dư=\dfrac{19,6}{419,5}\times100\%=4,67\%\)
\(C\%_{ZnSO_4}=\dfrac{48,3}{419,5}\times100\%=11,51\%\)
PTHH: Zn + H2SO4 → ZnSO4 + H2
\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
a) Theo PT: \(n_{Zn}pư=n_{H_2}=0,3\left(mol\right)\)
\(\Rightarrow m_{Zn}pư=0,3\times65=19,5\left(g\right)\)
b) \(m_{H_2SO_4}=400\times12,25\%=49\left(g\right)\)
\(\Rightarrow n_{H_2SO_4}=\dfrac{49}{98}=0,5\left(mol\right)\)
Theo PT: \(n_{H_2SO_4}=n_{H_2}\)
Theo bài: \(n_{H_2SO_4}=0,5>n_{H_2}=0,3\)
⇒ dd H2SO4 dư
Vậy dung dịch sau phản ứng gồm: H2SO4 dư và ZnSO4
Theo PT: \(n_{H_2SO_4}pư=n_{H_2}=0,3\left(mol\right)\)
\(\Rightarrow n_{H_2SO_4}dư=0,5-0,3=0,2\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4}dư=0,2\times98=19,6\left(g\right)\)
Theo PT: \(n_{ZnSO_4}=n_{H_2}=0,3\left(mol\right)\)
\(\Rightarrow m_{ZnSO_4}=0,3\times161=48,3\left(g\right)\)
\(\Sigma m_{dd}=19,5+400=419,5\left(g\right)\)
\(\Rightarrow C\%_{H_2SO_4}dư=\dfrac{19,6}{419,5}\times100\%=4,67\%\)
\(C\%_{ZnSO_4}=\dfrac{48,3}{419,5}\times100\%=11,51\%\)
