Mg + H2SO4 → MgSO4 + H2
\(n_{Mg}=\frac{3,6}{24}=0,15\left(mol\right)\)
\(m_{H_2SO_4}=200\times19,6\%=39,2\left(g\right)\)
\(\Rightarrow n_{H_2SO_4}=\frac{39,2}{98}=0,4\left(mol\right)\)
Theo Pt: \(n_{Mg}=n_{H_2SO_4}\)
Theo bài: \(n_{Mg}=\frac{3}{8}n_{H_2SO_4}\)
Vì \(\frac{3}{8}< 1\) ⇒ H2SO4 dư
a) Theo PT: \(n_{H_2}=n_{Mg}=0,15\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,15\times22,4=3,36\left(l\right)\)
b) \(m_{H_2}=0,15\times2=0,3\left(mol\right)\)
Ta có: \(m_{dd}saupư=3,6+200-0,3=203,3\left(g\right)\)
Theo PT: \(n_{H_2SO_4}pư=n_{Mg}=0,15\left(mol\right)\)
\(\Rightarrow n_{H_2SO_4}dư=0,4-0,15=0,25\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4}dư=0,25\times98=24,5\left(g\right)\)
\(\Rightarrow C\%_{H_2SO_4}dư=\frac{24,5}{203,3}\times100\%=12,05\%\)
Theo pT: \(n_{MgSO_4}=n_{Mg}=0,15\left(mol\right)\)
\(\Rightarrow m_{MgSO_4}=0,15\times120=18\left(g\right)\)
\(\Rightarrow C\%_{MgSO_4}=\frac{18}{203,3}\times100\%=8,85\%\)
